WEAK CHOICE - another query.

[Bill Taylor]

Along the lines of some recent threads.

QUESTION: Does either of (a) or (b) imply the other?

(a) A countable union of countable sets is countable.
(b) Every infinite set has a countable subset.

-------------------------------------------------------------------------------
Bill Taylor W.Ta...@math.canterbury.ac.nz
-------------------------------------------------------------------------------
Smoking is one of the leading causes of statistics.
-------------------------------------------------------------------------------

[Fred Galvin]

On 25 Aug 1997, ilias kastanas 08-14-90 wrote:

> Everyone knows the kind of dedication to AC you have, Bill; you
> cannot say too many good things about that Axiom. So it is obvious you
> know the _real_ answer, i.e. "Yes, trivially so... since both are true"!

That reminds me of an old Tarski anecdote, which I can barely recollect;
maybe somebody who remembers it better can correct me on the details.
It seems that, after he proved that the assertion "every infinite cardinal
is equal to its square" implies the axiom of choice, Tarski sent the paper
to Lebesgue (?) to communicate to the Paris Academy of Sciences (?).
Lebesgue replied saying he was very sorry, but he thinks the axiom of
choice is nonsense, and cannot in good conscience recommend a paper
proving that one nonsense implies another; but he will give the paper to
his colleague Hadamard (?) who believes in the axiom. The next letter
from Lebesgue was even more apologetic: Hadamard says the axiom of choice
is trivial, and he can't recommend a paper proving that one triviality
implies another.

[Ilias Kastanas]

In article <Pine.SOL.3.96.970829185329.26910A-100000@titania>,

Hmm... "... but don't worry, Mr. Tarski, we'll pass on your
paper to Poincare..." ?!

Did Tarski point out that "a^2 = b^2 -> a = b" is equivalent
to AC, while "2a = 2b -> a = b" holds regardless of AC? Trivialities
can relate non-trivially!

Anyway, Lebesgue would not be receptive to a formal point of view
given his "constructivism", in a generalized sense. (Story: At the
Scottish Cafe he was handed a menu written in Polish; he took a look and
gave it back saying "I'm sorry, I only eat definable things"). Sets and
functions exist if they are specifiable, somehow. It would have been
neat if matters "closed off" with Borel sets! Well, Borel images turned
out non-Borel, and the projective sets appeared by right; still, coanalytic
sets, say, have the uniformization property and there is no need to invoke
AC to make choices.

A "moral victory" for Lebesgue: we know today that in the absence
of AC it is consistent that all sets of reals are Lebesgue measurable. It
takes something non-constructive to produce a non-measurable set (although
not full AC; a nonprincipal ultrafilter on w is enough).

But as Lusin mentions, most reals are "undefinable"... and yet one
forms with them sets that _are_ definable. Worse, going up in the projective
hierarchy it seems definability all of a sudden doesn't help any more; after
just a couple of levels one cannot prove basic properties... nor refute them.
Ironically enough, one does hit formal, axiomatic issues; those properties
are in fact independent of set theory.

To push on the study of definable sets, one usually assumes Determi-
nacy (of games on them). Many "nice" properties of Borel and related sets
(say, the Galvin-Prikry theorem!) follow formally from Borel Determinacy
(itself provable in set theory). One lifts this situation to higher levels.

So there is some long-term consolation for Tarski!

Ilias

P.S. Eh, my very first sentence quoted
above should be read carefully, espe-
cially if you know Bill! Consider
it literally...

[Herman Rubin]

In article <5u8fmd$3...@gap.cco.caltech.edu>,

Ilias Kastanas <ika...@alumni.caltech.edu> wrote:
>In article <Pine.SOL.3.96.970829185329.26910A-100000@titania>,
>Fred Galvin <gal...@math.ukans.edu> wrote:
>>On 25 Aug 1997, ilias kastanas 08-14-90 wrote:

>>> Everyone knows the kind of dedication to AC you have, Bill; you
>>> cannot say too many good things about that Axiom. So it is obvious you
>>> know the _real_ answer, i.e. "Yes, trivially so... since both are true"!

>>That reminds me of an old Tarski anecdote, which I can barely recollect;
>>maybe somebody who remembers it better can correct me on the details.
>>It seems that, after he proved that the assertion "every infinite cardinal
>>is equal to its square" implies the axiom of choice, Tarski sent the paper
>>to Lebesgue (?) to communicate to the Paris Academy of Sciences (?).

This needs correction; let me give the version as I recall it from
Tarski himself. The correspondence was destroyed during WWII.

Sierpinski sent Tarski's paper to Lebesgue to communicated it to the
French Academy of Sciences. After a couple of weeks, he received a
letter from Lebesgue stating that the proofs looked correct, but as
he considered the axiom of choice to be nonsense, how could he submit
a paper to the French academy showing something is equivalent to
nonsense? But he was sending the paper to their mutual friend Hadamard,
who believes in the axiom of choice.

After a few weeks, he then received a letter from Hadamard, together
with the manuscript, stating that the proofs looked correct, but as
he considered the axiom of choice to be obvious, how could he submit
a paper to the French academy showing something is equivalent to the
obvious? The paper was then printed in _Fundamenta Mathematicae_,
of which Sierpinski was the editor.

.....................

> Did Tarski point out that "a^2 = b^2 -> a = b" is equivalent
>to AC, while "2a = 2b -> a = b" holds regardless of AC? Trivialities
>can relate non-trivially!

He could hardly state the latter, as the status of 2a = 2b -> a = b
was not known at the time. But 2a = 2b does not imply a = b without
some weak form of the axiom of choice.

.......................

> A "moral victory" for Lebesgue: we know today that in the absence
>of AC it is consistent that all sets of reals are Lebesgue measurable. It
>takes something non-constructive to produce a non-measurable set (although
>not full AC; a nonprincipal ultrafilter on w is enough).

We do not KNOW it is consistent; the present status is that this seems
to need the existence of inaccessible cardinals, which cannot be proved
consistent. Even the axiom of choice for a given collection of pairs
of subsets of reals is enough to get non-measurable sets.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558

[Matthew P Wiener]

In article <5ubro3$h...@mean.stat.purdue.edu>, hrubin@mean (Herman Rubin) writes:
>In article <5u8fmd$3...@gap.cco.caltech.edu>,
>Ilias Kastanas <ika...@alumni.caltech.edu> wrote:

>> A "moral victory" for Lebesgue: we know today that in the absence
>>of AC it is consistent that all sets of reals are Lebesgue measurable. It
>>takes something non-constructive to produce a non-measurable set (although
>>not full AC; a nonprincipal ultrafilter on w is enough).

>We do not KNOW it is consistent; the present status is that this seems

>to need the existence of inaccessible cardinals, [...]

The correct statement is "this is known to need the consistency of
inaccessible cardinals with Zermelo-Fraenkel set theory". There is
no need for the inaccessibles to actually exist, as one could cut
your von Neumann universe off at the first one. And the "is known"
aspect, as opposed to the "seems to need", refers to Shelah's proof
that Solovay's inaccessible was necessary. (Although bizarrely
enough, Shelah showed it was _not_ necessary for the Baire property.)
--
-Matthew P Wiener (wee...@sagi.wistar.upenn.edu)

[Ilias Kastanas]

In article <5ubu61$dhj$1...@netnews.upenn.edu>,

That was the "going to Canossa" announcement, wasn't it? At
first he thought measure and category went the same way.

Anyway, regarding the "correction", I'll repeat: we do KNOW it
is consistent.

No, I didn't "forget" the inaccessible; that's hardly likely, as
I posted a detailed explanation of Solovay's model, here or in sci.logic,
a few months ago. The point is, if I'm stating results formally, I'll
say "if PA is consistent", or whatever it takes. Here, however, I was
talking about what we know. Yes, we know that PA is consistent. In fact,
ZFC is consistent. For that matter, ZFC + EI (I inaccessible) is consi-
stent. And that comes out not by formal manipulations of symbols but
through our understanding of V.

Ilias

[Bill Taylor]

ika...@alumni.caltech.edu (Ilias Kastanas) writes:

|> >Lebesgue replied saying he was very sorry, but he thinks the axiom of
|> >choice is nonsense, and cannot in good conscience recommend a paper
|> >proving that one nonsense implies another;

<snigger> Nice chap, Lebesgue!

|> > Hadamard says the axiom of choice is trivial,

No doubt that was the time he said "How trivial! Oh, wait a minute..." and
half an hour later stormed back to his office grumping "I was right, it
*was* trivial after all!"

|> Did Tarski point out that "a^2 = b^2 -> a = b" is equivalent
|> to AC, while "2a = 2b -> a = b" holds regardless of AC? Trivialities
|> can relate non-trivially!

Amusing.

|> A "moral victory" for Lebesgue: we know today that in the absence
|> of AC it is consistent that all sets of reals are Lebesgue measurable.

I wish our measure theory lecturer had taken this view. We spent the whole
year with theorem after theorem beginning "For a measurable set S..." or
"For a measurable function f...", and there was never any other sort!

|> not full AC; a nonprincipal ultrafilter on w is enough

AHA! An unprincipled infiltrator!

|> >> Everyone knows the kind of dedication to AC you have, Bill; you
|> >> cannot say too many good things about that Axiom.

... ...

|> P.S. Eh, my very first sentence quoted
|> above should be read carefully, espe-
|> cially if you know Bill! Consider
|> it literally...

He heh. "I just can't *say* how much I enjoyed your party!"

Or best of all: "That's uncommonly generous of you!"

-------------------------------------------------------------------------------
Bill Taylor W.Ta...@math.canterbury.ac.nz
-------------------------------------------------------------------------------

Cynicism can be cured - but who cares - it wont help.
-------------------------------------------------------------------------------

[Robert Hill]

In article <5udc19$q1o$1...@cantuc.canterbury.ac.nz>, mat...@math.canterbury.ac.nz (Bill Taylor) writes:

> |> > Hadamard says the axiom of choice is trivial,
>
> No doubt that was the time he said "How trivial! Oh, wait a minute..." and
> half an hour later stormed back to his office grumping "I was right, it
> *was* trivial after all!"

I've heard that story many times, but never before with a name attached.
I'm glad to learn it was Hadamard. But in the versions I heard, it was
not in an office but in a lecture room, in the middle of a lecture, when
he introduced a step in the argument by "It is obvious that ...", and then
had second thoughts, and took half an hour to decide that it was obvious
after all.

--
Robert Hill

University Computing Service, Leeds University, England

"Though all my wares be trash, the heart is true."
- John Dowland, Fine Knacks for Ladies (1600)

[Ilias Kastanas]

In article <5udc19$q1o$1...@cantuc.canterbury.ac.nz>,
Bill Taylor <mat...@math.canterbury.ac.nz> wrote:
>ika...@alumni.caltech.edu (Ilias Kastanas) writes:
>

>|> > Hadamard says the axiom of choice is trivial,
>
>No doubt that was the time he said "How trivial! Oh, wait a minute..." and
>half an hour later stormed back to his office grumping "I was right, it
>*was* trivial after all!"

Eh... isn't it?

>|> Did Tarski point out that "a^2 = b^2 -> a = b" is equivalent
>|> to AC, while "2a = 2b -> a = b" holds regardless of AC? Trivialities
>|> can relate non-trivially!
>
>Amusing.

Pity he didn't mention the infinite set without a countable subset,
it would have tickled Lebesgue pink.

>|> not full AC; a nonprincipal ultrafilter on w is enough
>
>AHA! An unprincipled infiltrator!

All right, out goes the ultrafilter, and we bring in... AC-2 !
Choice for pairs. Now that should be OK...

>|> >> Everyone knows the kind of dedication to AC you have, Bill; you
>|> >> cannot say too many good things about that Axiom.
>... ...
>|> P.S. Eh, my very first sentence quoted
>|> above should be read carefully, espe-
>|> cially if you know Bill! Consider
>|> it literally...
>
>He heh. "I just can't *say* how much I enjoyed your party!"
>
>Or best of all: "That's uncommonly generous of you!"

But we know that deep in your heart, Bill, you _love_ the axiom...

Ilias

[Matthew P Wiener]

In article <5uc9e7$j...@gap.cco.caltech.edu>, ikastan@alumni (Ilias Kastanas) writes:
>In article <5ubu61$dhj$1...@netnews.upenn.edu>,
>Matthew P Wiener <wee...@sagi.wistar.upenn.edu> wrote:

>>>We do not KNOW it is consistent; the present status is that this seems
>>>to need the existence of inaccessible cardinals, [...]

>>The correct statement is "this is known to need the consistency of

>>inaccessible cardinals with Zermelo-Fraenkel set theory". [...]

> That was the "going to Canossa" announcement, wasn't it? At
>first he thought measure and category went the same way.

Yes. Is that his only major public blooper? He did catch himself,
so it hardly counts.

> Anyway, regarding the "correction", I'll repeat: we do KNOW it

>is consistent. [...]

I was correcting Herman Rubin's "present status" statement of fact, not
the epistemological issue.

[Ilias Kastanas]

In article <5ueott$b21$1...@netnews.upenn.edu>,

Matthew P Wiener <wee...@sagi.wistar.upenn.edu> wrote:
>In article <5uc9e7$j...@gap.cco.caltech.edu>, ikastan@alumni (Ilias Kastanas) writes:
>>In article <5ubu61$dhj$1...@netnews.upenn.edu>,
>>Matthew P Wiener <wee...@sagi.wistar.upenn.edu> wrote:
>
>>>>We do not KNOW it is consistent; the present status is that this seems
>>>>to need the existence of inaccessible cardinals, [...]
>
>>>The correct statement is "this is known to need the consistency of
>>>inaccessible cardinals with Zermelo-Fraenkel set theory". [...]
>
>> That was the "going to Canossa" announcement, wasn't it? At
>>first he thought measure and category went the same way.

>Yes. Is that his only major public blooper? He did catch himself,
>so it hardly counts.

I have never thought of it as an adverse count. He seems willing
to talk about work in progress anyway, things he finds plausible but has
no proof yet, and so on; which is fine, and very interesting too.

>> Anyway, regarding the "correction", I'll repeat: we do KNOW it
>>is consistent. [...]
>
>I was correcting Herman Rubin's "present status" statement of fact, not
>the epistemological issue.

I didn't mean your comments, sorry if I wasn't clear. If memory
serves, you wouldn't disagree anyway.

Ilias

[ilias kastanas 08-14-90]

In article <5ubro3$h...@mean.stat.purdue.edu>,
Herman Rubin <hru...@mean.stat.purdue.edu> wrote:
@In article <5u8fmd$3...@gap.cco.caltech.edu>,
@Ilias Kastanas <ika...@alumni.caltech.edu> wrote:
@>In article <Pine.SOL.3.96.970829185329.26910A-100000@titania>,
@>Fred Galvin <gal...@math.ukans.edu> wrote:
@>>On 25 Aug 1997, ilias kastanas 08-14-90 wrote:
@
@>>> Everyone knows the kind of dedication to AC you have, Bill; you
@>>> cannot say too many good things about that Axiom. So it is obvious you
@>>> know the _real_ answer, i.e. "Yes, trivially so... since both are true"!
@
@>>That reminds me of an old Tarski anecdote, which I can barely recollect;
@>>maybe somebody who remembers it better can correct me on the details.
@>>It seems that, after he proved that the assertion "every infinite cardinal
@>>is equal to its square" implies the axiom of choice, Tarski sent the paper
@>>to Lebesgue (?) to communicate to the Paris Academy of Sciences (?).
@
@This needs correction; let me give the version as I recall it from
@Tarski himself. The correspondence was destroyed during WWII.
@
@Sierpinski sent Tarski's paper to Lebesgue to communicated it to the
@French Academy of Sciences. After a couple of weeks, he received a
@letter from Lebesgue stating that the proofs looked correct, but as
@he considered the axiom of choice to be nonsense, how could he submit
@a paper to the French academy showing something is equivalent to
@nonsense? But he was sending the paper to their mutual friend Hadamard,
@who believes in the axiom of choice.
@
@After a few weeks, he then received a letter from Hadamard, together
@with the manuscript, stating that the proofs looked correct, but as
@he considered the axiom of choice to be obvious, how could he submit
@a paper to the French academy showing something is equivalent to the
@obvious? The paper was then printed in _Fundamenta Mathematicae_,
@of which Sierpinski was the editor.
@
@ .....................
@
@> Did Tarski point out that "a^2 = b^2 -> a = b" is equivalent
@>to AC, while "2a = 2b -> a = b" holds regardless of AC? Trivialities
@>can relate non-trivially!

@He could hardly state the latter, as the status of 2a = 2b -> a = b
@was not known at the time.

He could have, since the result was in fact known.

"2a = 2b -> a = b" is the subject of a paper by Sierpinski, publi-
shed in Fundamenta Math., in 1922. It goes back to Bernstein, Math. Ann.
1905.

Tarski's paper appeared in 1924. So it is natural, possible and
plausible that Tarski did know the result at the pertinent time.

If Tarski has told you he didn't know, that's different and settles
the matter. If not, I see no reason to assume it.

@was not known at the time. But 2a = 2b does not imply a = b without
@some weak form of the axiom of choice.

No, that is not right.

One can prove a = b from 2a = 2b by an explicit construction
of an 1-1 onto mapping. No form of the axiom of choice is involved.

And the fact that a posting of mine is jocular and light-hearted
does not imply its mathematical content is careless and loose!

.......................
@
@> A "moral victory" for Lebesgue: we know today that in the absence
@>of AC it is consistent that all sets of reals are Lebesgue measurable. It
@>takes something non-constructive to produce a non-measurable set (although
@>not full AC; a nonprincipal ultrafilter on w is enough).
@
@We do not KNOW it is consistent; the present status is that this seems
@to need the existence of inaccessible cardinals, which cannot be proved
@consistent.

But, as posted elsewhere, we do know it is consistent.

@consistent. Even the axiom of choice for a given collection of pairs
@of subsets of reals is enough to get non-measurable sets.

Eh, what do you mean by "even"? My assumption, a nonprincipal
ultrafilter on w, is weaker than AC for pairs -- and of course much
weaker than the Boolean Prime Ideal principle.

Ilias

[Matthew P Wiener]

In article <5uih0j$bme$1...@cantuc.canterbury.ac.nz>, mathwft@math (Bill Taylor) writes:
>Well actually, I *do*. You might not have been on the net when I had
>some good things to say about it. Mainly, that, now that we know it's
>ZF-independent, it provides a valuable warning against trying to construct
>examples that are equivalent to it (or to a weaker form). Also anything
>proved using it, cannot be disproved otherwise. All very useful.
>Though the same could be said of many other axioms for the same reason.

You may be happy to learn that Shelah has roughly the same attitude to
independence results (but he includes choice in his beliefs). In particular,
he believes Real Mathematics is what you can prove in ZFC, and that the Real
Value of independence proofs is more as a hunting ground and warning zone.

[Andrew Boucher]

In article <5uih0j$bme$1...@cantuc.canterbury.ac.nz>,
mat...@math.canterbury.ac.nz (Bill Taylor) wrote:

>All that's wrong with it (Axiom of Choice), is that it's false. :) And not
even false in set
>theory, as the "standard model" V is so non-"real" it doesn't matter much
>*what* you consistently assume about it. It's just that it's false in
>analysis, and importantly so when used to prove existential results.
>(As for universal results - who cares? What matter if every vector space
>has a basis, when any particular one you ever work with is certain to
>have one anyway...)

Which approach is preferable:
1) State a theorem generalizing over all vector spaces and in the first
line of proof assert by AC the existence of a basis; or
2) State the theorem generalizing over all vector spaces with bases and
not use AC?

Surely 2 is better. As you say any particular one we work with has a basis
in any case, so the extra generality that 1 brings is not useful--and it
comes at the price of assuming an axiom which is questionable.

[Fred Galvin]

On 31 Aug 1997, Herman Rubin wrote:

> >Fred Galvin <gal...@math.ukans.edu> wrote:
> >>That reminds me of an old Tarski anecdote, which I can barely recollect;

> >>maybe somebody who remembers it better can correct me on the details.

[. . .]

> This needs correction; let me give the version as I recall it from

> Tarski himself. The correspondence was destroyed during WWII.

Many thanks for the corrections! That's why I posted the anecdote; I hoped
somebody with a better memory would post a more accurate version. I got it
from the horse's mouth too, but my memory isn't that good. I'm glad that
at least I got the names of the two "villains" right. I wonder if the
story was published anywhere?

> Sierpinski sent Tarski's paper to Lebesgue to communicated it to the

> French Academy of Sciences. After a couple of weeks, he received a

> letter from Lebesgue stating that the proofs looked correct, but as

> he considered the axiom of choice to be nonsense, how could he submit

> a paper to the French academy showing something is equivalent to

> nonsense? But he was sending the paper to their mutual friend Hadamard,

> who believes in the axiom of choice.
>

> After a few weeks, he then received a letter from Hadamard, together

> with the manuscript, stating that the proofs looked correct, but as

> he considered the axiom of choice to be obvious, how could he submit

> a paper to the French academy showing something is equivalent to the

> obvious? The paper was then printed in _Fundamenta Mathematicae_,

[Fred Galvin]

On 31 Aug 1997, Herman Rubin wrote:

> In article <5u8fmd$3...@gap.cco.caltech.edu>,
> Ilias Kastanas <ika...@alumni.caltech.edu> wrote:

> >Did Tarski point out that "a^2 = b^2 -> a = b" is equivalent

> >to AC, while "2a = 2b -> a = b" holds regardless of AC? Trivialities

> >can relate non-trivially!

>
> He could hardly state the latter, as the status of 2a = 2b -> a = b

> was not known at the time.

Are you sure about that? I can't go to the library now to check the
original articles, but I think Sierpinski is a pretty reliable source. On
p. 174 of _Cardinal and Ordinal Numbers_, 2nd edition, 1965, he writes:
"We are able to prove without the aid of the axiom of choice a more
general theorem . . . equivalent to the following one: If, for a natural
number k and cardinal numbers m and n, we have km = kn, then m = n. This
theorem was proved without the aid of the axiom of choice for k = 2 in
1905 [two decades before Tarski's first paper] by F. Bernstein, who also
outlined a proof for natural k . . ." The reference is F. Bernstein,
Untersuchungen aus der Mengenlehre, Math. Annalen 61 (1905), p. 117-155. A
detailed proof for all finite k appeared in A. Tarski, Cancellation laws
in the arithmetic of cardinals, Fund. Math. 36 (1949), 77-92.

> But 2a = 2b does not imply a = b without

> some weak form of the axiom of choice.

Are you sure about that? Sierpinski says "without the aid of the axiom of
choice". I recall working out the proof as an exercise (some 30 years ago)
and it seemed O.K. at the time, but the details are kind of hazy now. I'll
have to think about it again.

[Fred Galvin]

O.K., I remember how to prove that 2a = 2b implies a = b; the proof does
not use any choice at all, and is rather trivial. Maybe you were thinking
of some other proposition which requires choice?

[Ilias Kastanas]

In article <aboucher-060...@cyber133-3.paris.imaginet.fr>,

Andrew Boucher <abou...@imagiNet.fr> wrote:
>In article <5uih0j$bme$1...@cantuc.canterbury.ac.nz>,
>mat...@math.canterbury.ac.nz (Bill Taylor) wrote:
>
>>All that's wrong with it (Axiom of Choice), is that it's false. :) And not
>even false in set
>>theory, as the "standard model" V is so non-"real" it doesn't matter much
>>*what* you consistently assume about it. It's just that it's false in
>>analysis, and importantly so when used to prove existential results.
>>(As for universal results - who cares? What matter if every vector space
>>has a basis, when any particular one you ever work with is certain to
>>have one anyway...)

>Which approach is preferable:
> 1) State a theorem generalizing over all vector spaces and in the first
>line of proof assert by AC the existence of a basis; or
> 2) State the theorem generalizing over all vector spaces with bases and
>not use AC?
>
>Surely 2 is better.

I hope you were careful in doing that proof, e.g. in using dimension.
Without AC, there can be a vector space that has two bases of different
cardinalities.

>Surely 2 is better. As you say any particular one we work with has a basis
>in any case, so the extra generality that 1 brings is not useful--and it

I've mentioned this before: is it "evident" that R has a basis over Q?

>comes at the price of assuming an axiom which is questionable.

Maybe it's worth discussing on what grounds might AC be questionable.
Not having been here too long, I don't recall any such debate. (I hope we
won't be disturbing threads about a trillion Bible codes forming ley lines
on the Great Pyramid in 2001).

So then, are there cogent objections to AC? And do they apply to
just AC_w too? (Choice for countably many sets, E_0, E_1, ...).

Ilias

[Fred Galvin]

On 2 Sep 1997, ilias kastanas 08-14-90 wrote:

> One can prove a = b from 2a = 2b by an explicit construction

> of an 1-1 onto mapping. No form of the axiom of choice is involved.

Yes! Here's how I remember that construction. Suppose we have a white
horses and b black horses. (Someone was asking about horses on another
thread.) Then the set comprising the heads and tails of the white horses
has cardinality 2a; likewise the set of heads and tails of black horses
has cardinality 2b. By hypothesis we have a bijection between these two
sets. Each "component" or "orbit" of the bijection is either a circle or
an endless line of horses, alternating in color, each horse facing one of
its two neighbors. It will suffice to show how to pair off white and black
horses in a single orbit in a choice-free way. Consider the case of an
infinite orbit (the finite case is the same but simpler).

Step 1. Pair off any horses that face each other. Remove all such pairs
and close up the gaps. (In case an infinite number of horses got paired,
the new line may still be endless, or it may have one or two endpoints;
that doesn't matter. On the other hand, it's possible that no horses got
paired in this step.)

Step 2. Turn all the white horses 180 degrees, remove any facing pairs,
and close up the gaps.

Step 3. Turn the black horses around, etc.

Step 4. Turn the white horses around, etc. (By this time, some horses will
surely have been paired off.)

Keep doing this until either all horses have been paired off, or else you
are left with just one horse. In the latter case, put all the horses back
in their starting positions. Calling the leftover horse "Zero" and the
horse it's facing "One", define a bijection between the set of horses and
the set of integers, and pair horse 2n with horse 2n+1.

The procedure for a finite orbit is exactly the same, except that we can't
have a horse left over.

[Matthew P Wiener]

In article <aboucher-060...@cyber133-3.paris.imaginet.fr>, "Andrew Boucher" <aboucher@imagiNet writes:
>Which approach is preferable:

> 1) State a theorem generalizing over all vector spaces and in
>the first line of proof assert by AC the existence of a basis; or

> 2) State the theorem generalizing over all vector spaces with
>bases and not use AC?

>Surely 2 is better. As you say any particular one we work with has a

>basis in any case, so the extra generality that 1 brings is not

>useful--and it comes at the price of assuming an axiom which is
>questionable.

My favorite moment in the choice wars was in an introductory several
complex variables course by Robert Gunning. In attendance was one
pesky fellow, repeatedly (as in several times per lecture, usually
once every two or three minutes) asking questions that seemed to serve
only to side track onto details the pesky fellow was proud to know
about. Gunning handled him rather well, but one time was precious.

(Since Gunning at the time was department chairman, and was revising
his series of lecture note books on the subject, while the pesky fellow
was not a student, but an untenured up-and-coming hopeful who did not
think he needed to prove anything to anybody, it may well have been a
deliberate arrangement. But he sure *seemed* pesky!)

Anwyay, there's a basic fact that certain relevant sheaves are flabby
(or something along those lines), and the proof that Gunning gave
involved the axiom of choice. As was known to everyone in the class,
one can give a choice-free proof, if the base space was paracompact.
And of course, the pesky fellow had to bring this up. "Excuse me, but
couldn't you prove ... without using the axiom of choice, if the base
space is paracompact?!"

Gunning instantly replied "You can assume paracompactness, you can
assume the axiom of choice, you can assume whatever you want!!" That
cracked everyone else up, and the pesky fellow was rather quiet the
rest of that particular lecture.

[Ilias Kastanas]

In article <Pine.SOL.3.96.970907131732.24156A-100000@titania>,

Fred Galvin <gal...@math.ukans.edu> wrote:
>
>On 2 Sep 1997, ilias kastanas 08-14-90 wrote:
>
>> One can prove a = b from 2a = 2b by an explicit construction
>> of an 1-1 onto mapping. No form of the axiom of choice is involved.

>Yes! Here's how I remember that construction. Suppose we have a white
>horses and b black horses. (Someone was asking about horses on another

Aha... Well, I had in mind a very, very different proof.
No horses. Knives and spoons.

This proof obviously uses EC. (Equestrian Choice).

>thread.) Then the set comprising the heads and tails of the white horses
>has cardinality 2a; likewise the set of heads and tails of black horses
>has cardinality 2b. By hypothesis we have a bijection between these two
>sets. Each "component" or "orbit" of the bijection is either a circle or
>an endless line of horses, alternating in color, each horse facing one of
>its two neighbors. It will suffice to show how to pair off white and black
>horses in a single orbit in a choice-free way. Consider the case of an
>infinite orbit (the finite case is the same but simpler).
>
>Step 1. Pair off any horses that face each other. Remove all such pairs
>and close up the gaps. (In case an infinite number of horses got paired,
>the new line may still be endless, or it may have one or two endpoints;
>that doesn't matter. On the other hand, it's possible that no horses got
>paired in this step.)
>
>Step 2. Turn all the white horses 180 degrees, remove any facing pairs,
>and close up the gaps.

It sure is much more vivid than two-way infinite sequences, and
iterating bijections!

>Step 3. Turn the black horses around, etc.
>
>Step 4. Turn the white horses around, etc. (By this time, some horses will
>surely have been paired off.)
>
>Keep doing this until either all horses have been paired off, or else you
>are left with just one horse. In the latter case, put all the horses back
>in their starting positions. Calling the leftover horse "Zero" and the

All this was just to get a "starting horse"!?... Meanwhile, the
chap who had been screaming "My Kingdom for a horse!..." has gone hoarse.

The proof method also applies to some shopping excursions.

>horse it's facing "One", define a bijection between the set of horses and
>the set of integers, and pair horse 2n with horse 2n+1.
>
>The procedure for a finite orbit is exactly the same, except that we can't
>have a horse left over.

Yes, the same directions apply, which is convenient;... because,
if I may point out the obvious, we surely may not handle first this orbit,
then that one.... any more than we may pick first this pair, then the
other! So we proceed with all orbits at once; just like we remove from
an orbit all appropriate pairs as a single operation.

Who was it who said, "You can do whatever you like, as long as
you do not frighten the horses"?

Ilias

[Herman Rubin]

In article <5usfj2$4...@gap.cco.caltech.edu>,

Ilias Kastanas <ika...@alumni.caltech.edu> wrote:
>In article <aboucher-060...@cyber133-3.paris.imaginet.fr>,

>Andrew Boucher <abou...@imagiNet.fr> wrote:
>>In article <5uih0j$bme$1...@cantuc.canterbury.ac.nz>,
>>mat...@math.canterbury.ac.nz (Bill Taylor) wrote:

....................

> I hope you were careful in doing that proof, e.g. in using dimension.
>Without AC, there can be a vector space that has two bases of different
>cardinalities.

>>Surely 2 is better. As you say any particular one we work with has a basis
>>in any case, so the extra generality that 1 brings is not useful--and it

> I've mentioned this before: is it "evident" that R has a basis over Q?

>>comes at the price of assuming an axiom which is questionable.

> Maybe it's worth discussing on what grounds might AC be questionable.
>Not having been here too long, I don't recall any such debate. (I hope we
>won't be disturbing threads about a trillion Bible codes forming ley lines
>on the Great Pyramid in 2001).

> So then, are there cogent objections to AC? And do they apply to
>just AC_w too? (Choice for countably many sets, E_0, E_1, ...).

There are lots of cogent objections. The Banach-Mazur game, and other
games of this type, which do not have values; the Banach-Tarski paradox;
the existence of non-measurable sets (although this could be argued
either way); and many others.

For example, the existence of a basis of R over Q immediately produces
a countable number of sets whose union is the unit interval.

On the other hand, the axiom of choice simplifies many situations.
Much of point-set topology depends heavily on it, or at least on the
Prime Ideal Theorem, which is strictly weaker.

For analysis, choice for countably many sets is not quite what is needed,
as one needs dependent choice. This is a sequence of choices each
depending on the previous ones.

[Fred Galvin]

On 7 Sep 1997, Ilias Kastanas wrote:

> It sure is much more vivid than two-way infinite sequences, and
> iterating bijections!

Maybe, but it's also a hell of a lot easier to type.

> All this was just to get a "starting horse"!?...

I'm afraid so. Is there a neater way to do it? I'd like to see it.

> Yes, the same directions apply, which is convenient;... because,
> if I may point out the obvious, we surely may not handle first this orbit,
> then that one.... any more than we may pick first this pair, then the
> other! So we proceed with all orbits at once; just like we remove from
> an orbit all appropriate pairs as a single operation.

If I may also point out the obvious (for the sake of any beginning
students who may be reading this), there would be nothing wrong with
dividing into two cases, and handling the infinite orbits one way, the
finite orbits another. In fact, it's easy to do the finite-orbits case in
a way that doesn't generalize to infinite orbits. But there's no need for
that, since the procedure used for the infinite orbits also works for the
finite ones.

> Who was it who said, "You can do whatever you like, as long as
> you do not frighten the horses"?

According to Bartlett's (15th ed., p. 706) it was Mrs. Patrick Campbell
[Beatrice Stella Tanner Campbell], 1865-1940, who said "My dear, I don't
care what they do, so long as they don't do it in the street and frighten
the horses."

[Bill Taylor]

Fred Galvin <gal...@math.ukans.edu> does some dressage work...

|> Yes! Here's how I remember that construction. Suppose we have a white
|> horses and b black horses.

A most amusing proof! But let me see...

|> Then the set comprising the heads and tails of the white horses
|> has cardinality 2a;

Well, there's a subtle point here, isn't there? This gives set 2a as a+a.
i.e. 2 lots of a. I guess this is what 2a means? But it is not necessarily
equal to a2, is it, in the absence of choice? That (a2) would be a lots of 2,
i.e. "a" lots of unordered pairs, which couldn't automatically be classified
as heads and tails. (Who brought coin-tossing into this!) That is, instead
of a bunch of a black horses, we'd have a bunch of black push-me-pull-yous
whose ends can't be distinguished. Then I think the proof breaks down.

So that's a horse of a different color!

If I'm right, that means that: whereas 2a = 2b ==> a = b,
it is not so that a2 = b2 ==> a = b,

Have I got that right, anyone? Interesting.

Looking at Fred's tops-&-tails proof, it seems that "all" it boils down to is
noticing that 2a == 2b means there is an INJECTION of a into b, (just take
take half of the LHS and compress the RHS in the obvious way). And by symmetry
there is also an injection of b into a.

So the choice-free Schroder-Bernstein jumps this hurdle quite easily. There
was indeed a strong farmyard smell of SB (BS?) about Fred's nice proof anyway.

Hopefully that is all OK. Someone please tell me it is!

Well, guess I'll just saddle up and mosey off now...

-------------------------------------------------------------------------------
Bill Taylor W.Ta...@math.canterbury.ac.nz
-------------------------------------------------------------------------------

Salesman at night, farm girl's delight;
Salesman in morning, cattle take warning!
-------------------------------------------------------------------------------

[Brian M. Scott]

On 9 Sep 1997 04:41:10 GMT, mat...@math.canterbury.ac.nz (Bill
Taylor) wrote:

>Fred Galvin <gal...@math.ukans.edu> does some dressage work...

>|> Yes! Here's how I remember that construction. Suppose we have a white
>|> horses and b black horses.

>|> Then the set comprising the heads and tails of the white horses
>|> has cardinality 2a;

>Well, there's a subtle point here, isn't there? This gives set 2a as a+a.
>i.e. 2 lots of a.

I'd have said that it was a lots of 2: a horses, each with 2 ends.
But as long as we're doing cardinal arithmetic, there shouldn't be a
problem: given any two sets A and B, choice isn't needed to define a
bijection between A x B and B x A.

>I guess this is what 2a means?

For cardinal arithmetic it means |2 x a|. For ordinal arithmetic the
distinction matters, and 2a is usually taken to mean a lots of 2.
Writing w for omega, I understand w2 as w + w and 2w as w.

> But it is not necessarily
>equal to a2, is it, in the absence of choice? That (a2) would be a lots of 2,
>i.e. "a" lots of unordered pairs, which couldn't automatically be classified
>as heads and tails. (Who brought coin-tossing into this!) That is, instead
>of a bunch of a black horses, we'd have a bunch of black push-me-pull-yous
>whose ends can't be distinguished. Then I think the proof breaks down.

I believe that you're looking at a different theorem. As I understand
this one, it says that if f : 2 x a --> 2 x b is a bijection, then
there is a bijection between a and b. In this case the beasts are
definitely horses, not push-me-pull-yous: for horse x \in a, the head
is <0,x>, and the tail is <1,x>. I believe that you're thinking of a
result something like this:

Suppose that |A| = |B|, where A = U{P(x) : x \in a}, B = U{Q(x) : x
\in b}, each union is a disjoint union, and each P(x) and Q(x) is a
2-element set; then |a| = |b|.

Without choice for pairs I don't see how to adapt Fred's proof to this
result. I'm rusty, so I may be missing something obvious, but I don't
immediately see a choice-free argument of any kind. Without some
choice the converse of this version definitely fails: let a be a
countably infinite collection of pairs having no choice function, and
let b = {n x 2 : n \in w}. Clearly |a| = |b|, but a bijection between
A and B would induce a linear ordering of A and hence a choice
function for a. (So if we could get A to arrange itself along a
suitable ley line, we'd be in business: the head closer to the Great
Pyramid could be redefined as a tail! Sorry, Ilias.)

Brian M. Scott

[Ilias Kastanas]

In article <5v1ctv$1j...@mean.stat.purdue.edu>,

There is a progression of needs. AC_w suffices for basic analysis,
e.g. the usual development of Borel sets and Lebesgue measure. At some
stage one will invoke DC, maybe for Baire category in the absence of a
countable basis. Of course DC is strictly stronger than AC_w.

Considering AC_k and DC_k, where k is any aleph (DC = DC_w)
we have a source of stronger hypotheses. It turns out that "for all k, AC_k"
implies DC (but not DC_w_1), while just k <= some bound does not. DC
does not imply higher AC_k either. On the other hand "for all k DC_k"
wellorders every set and hence is full AC.

BPI has to come in for Stone Representation. It's the usual hypo-
thesis for Hahn-Banach (actually H-B is weaker than BPI). Or the Stone-Cech
compactification.

BPI is "to the side": it neither implies nor is implied by DC. (If
we add to a model M countably many Cohen-generic reals and generically a
set A that collects them, then A is infinite but Dedekind-finite in M and
hence AC_w and DC fail; but BPI holds. If we do not add A, BPI fails but
DC holds).

Tychonoff's theorem for 2^Y is equivalent to BPI. The full theorem
(or just for X^Y) of course amounts to full AC.

So, how high one will go depends on the subject matter!

Regarding the objections, they are related to each other. Banach-
Mazur and other games characterize regularity properties (measurability,
property of Baire, perfect set property...). Banach-Tarski decompositions
obviously involve non-measurable sets. So I think the "complaint" is: AC
implies that not all sets of reals are "regular".

I agree, this can be argued both ways! To what extent is it an
actual objection? Should it be that all sets of reals are regular -- on
principle? I have trouble seeing this as plausible -- in particular, more
plausible than "the Cartesian product of a family of nonempty sets is non-
empty"! The fact that this natural statement implies Tychonoff etc above,
provides bases and algebraic closures, makes every subgroup of a free group
free, and so on, _is_ remarkable, to be sure.

Banach-Tarski, by the way, is not quite a "perversity solely due
to AC". Statements of similar flavor, e.g. there exists a bounded set
congruent to a proper subset, hold _without_ AC, even on the plane. Also,
after seeing its proof, one usually doesn't consider B-T scandalous any
more. It is pretty much like Vitali's non-measurable set construction,
AC picking an element from each coset of the rationals.

I don't know how many people accept as valid objections to AC the
ones just discussed. And we have not heard yet from proponents of the
"fundamental" objection of non-effectiveness.

Ilias

[Ilias Kastanas]

In article <Pine.SOL.3.96.970908183247.20592C-100000@titania>,

Fred Galvin <gal...@math.ukans.edu> wrote:
>On 7 Sep 1997, Ilias Kastanas wrote:
>
>> It sure is much more vivid than two-way infinite sequences, and
>> iterating bijections!
>
>Maybe, but it's also a hell of a lot easier to type.
>
>> All this was just to get a "starting horse"!?...
>
>I'm afraid so. Is there a neater way to do it? I'd like to see it.

Oh, no; I was just voicing the complaint of all those poor
horses who have to gallop back in place for another round!

>> Yes, the same directions apply, which is convenient;... because,
>> if I may point out the obvious, we surely may not handle first this orbit,
>> then that one.... any more than we may pick first this pair, then the
>> other! So we proceed with all orbits at once; just like we remove from
>> an orbit all appropriate pairs as a single operation.
>
>If I may also point out the obvious (for the sake of any beginning
>students who may be reading this), there would be nothing wrong with
>dividing into two cases, and handling the infinite orbits one way, the
>finite orbits another. In fact, it's easy to do the finite-orbits case in
>a way that doesn't generalize to infinite orbits. But there's no need for
>that, since the procedure used for the infinite orbits also works for the
>finite ones.

Right, it is only a matter of convenience; what is a necessity is
having enough uniformity in the construction to avoid more than finitely
many choices.

Ilias

[Ilias Kastanas]

In article <3414df31...@news.csuohio.edu>,

Brian M. Scott <sc...@math.csuohio.edu> wrote:
>On 9 Sep 1997 04:41:10 GMT, mat...@math.canterbury.ac.nz (Bill
>Taylor) wrote:
>
>>Fred Galvin <gal...@math.ukans.edu> does some dressage work...
>
>>|> Yes! Here's how I remember that construction. Suppose we have a white
>>|> horses and b black horses.
>
>>|> Then the set comprising the heads and tails of the white horses
>>|> has cardinality 2a;
>
>>Well, there's a subtle point here, isn't there? This gives set 2a as a+a.
>>i.e. 2 lots of a.
>
>I'd have said that it was a lots of 2: a horses, each with 2 ends.
>But as long as we're doing cardinal arithmetic, there shouldn't be a
>problem: given any two sets A and B, choice isn't needed to define a
>bijection between A x B and B x A.

I understand 2a is intended as a+a here, implemented by sets
X, Y and a bijection between them. Of course we can then get a family
of ordered pairs indexed by X whose union is the same as X U Y. So we
do have either one of A x 2 or 2 x A.

What is _not_ intended is a-many _unordered_ pairs -- as you
mention below.

>>I guess this is what 2a means?
>
>For cardinal arithmetic it means |2 x a|. For ordinal arithmetic the
>distinction matters, and 2a is usually taken to mean a lots of 2.
>Writing w for omega, I understand w2 as w + w and 2w as w.
>
>> But it is not necessarily
>>equal to a2, is it, in the absence of choice? That (a2) would be a lots of 2,
>>i.e. "a" lots of unordered pairs, which couldn't automatically be classified
>>as heads and tails. (Who brought coin-tossing into this!) That is, instead
>>of a bunch of a black horses, we'd have a bunch of black push-me-pull-yous
>>whose ends can't be distinguished. Then I think the proof breaks down.
>
>I believe that you're looking at a different theorem. As I understand
>this one, it says that if f : 2 x a --> 2 x b is a bijection, then
>there is a bijection between a and b. In this case the beasts are
>definitely horses, not push-me-pull-yous: for horse x \in a, the head
>is <0,x>, and the tail is <1,x>. I believe that you're thinking of a
>result something like this:
>
>Suppose that |A| = |B|, where A = U{P(x) : x \in a}, B = U{Q(x) : x
>\in b}, each union is a disjoint union, and each P(x) and Q(x) is a
>2-element set; then |a| = |b|.
>
>Without choice for pairs I don't see how to adapt Fred's proof to this
>result. I'm rusty, so I may be missing something obvious, but I don't
>immediately see a choice-free argument of any kind. Without some

Worse: Without some kind of Choice, a2 meaning 2 + 2 + 2 ... (a
times) isn't even well-defined. It is supposed to be the "common cardina-
lity" of what we get when each (cardinal!) 2 stands for whatever 2-element
set (as long as all are disjoint). But such a common cardinality does
not necessarily exist, by your example of pairs not having a choice function.

>choice the converse of this version definitely fails: let a be a
>countably infinite collection of pairs having no choice function, and
>let b = {n x 2 : n \in w}. Clearly |a| = |b|, but a bijection between
>A and B would induce a linear ordering of A and hence a choice

I am sure you meant to capitalize a and b, and say that |w| = |w|
but |A| = B| fails.

Obviously, the reason for this mixup is that you are under Mummy's
curse -- for daring to insult the Great Pyramid below.

>function for a. (So if we could get A to arrange itself along a
>suitable ley line, we'd be in business: the head closer to the Great
>Pyramid could be redefined as a tail! Sorry, Ilias.)

Feh! I'm not worried. The Bible code says you cannot arrange
more than 2001 elements of A along a ley line, whatever units of measurement
you might employ; not in a trillion years.

Shantih shantih shantih

I.L.I.A.S.

(Inner Light Immanent After Satori)

[ilias kastanas 08-14-90]

In article <aboucher-090...@cyber124.paris.imaginet.fr>,
Andrew Boucher <abou...@imagiNet.fr> wrote:
@In article <5usfj2$4...@gap.cco.caltech.edu>, ika...@alumni.caltech.edu
@(Ilias Kastanas) wrote:
@
@>
@>In article <aboucher-060...@cyber133-3.paris.imaginet.fr>,
@>Andrew Boucher <abou...@imagiNet.fr> wrote:
@>>In article <5uih0j$bme$1...@cantuc.canterbury.ac.nz>,
@>>mat...@math.canterbury.ac.nz (Bill Taylor) wrote:
@>>
@>>>All that's wrong with it (Axiom of Choice), is that it's false. :) And
@not
@>>even false in set
@>>>theory, as the "standard model" V is so non-"real" it doesn't matter much
@>>>*what* you consistently assume about it. It's just that it's false in
@>>>analysis, and importantly so when used to prove existential results.
@>>>(As for universal results - who cares? What matter if every vector space
@>>>has a basis, when any particular one you ever work with is certain to
@>>>have one anyway...)
@>
@>>Which approach is preferable:
@>> 1) State a theorem generalizing over all vector spaces and in the
@first
@>>line of proof assert by AC the existence of a basis; or
@>> 2) State the theorem generalizing over all vector spaces with
@bases and
@>>not use AC?
@>>
@>>Surely 2 is better.
@>
@>
@> I hope you were careful in doing that proof, e.g. in using dimension.
@>Without AC, there can be a vector space that has two bases of different
@>cardinalities.
@>
@>
@>>Surely 2 is better. As you say any particular one we work with has a basis
@>>in any case, so the extra generality that 1 brings is not useful--and it
@>
@>
@> I've mentioned this before: is it "evident" that R has a basis over Q?
@>
@
@I myself didnÕt mention "evident"--perhaps I missed a message. What
@troubles me is a theorem which generalizes over all vector spaces and which
@really needs to generalize over all vector spaces with bases. This is done
@by invoking AC at the beginning of the proof. Since I find AC questionable,
@I donÕt like this.

You mentioned that "any particular space we work with has a basis
in any case". So I asked about R over Q. Is it one of the "vector
spaces with bases"? (And if yes, does it have dimension? ... c.f. above)

@>>comes at the price of assuming an axiom which is questionable.
@>
@>
@> Maybe it's worth discussing on what grounds might AC be
@questionable.
@>Not having been here too long, I don't recall any such debate. (I hope we
@>won't be disturbing threads about a trillion Bible codes forming ley lines
@>on the Great Pyramid in 2001).
@>
@> So then, are there cogent objections to AC?

@ArenÕt we getting this backwards? ArenÕt you the one who is assuming
@something? So shouldnÕt you be the one to provide reasons why I should
@accept it? Otherwise, I should be able to invoke OccamÕs Razor.

Uh, if you can provide reasons why I should accept Occam's Razor...

No kidding. It's not forwards or backwards; I don't _have_ to
defend it, and you don't _have_ to attack it. It's all voluntary.

@Is it intuitively obvious?

Yes. If sets are nonempty, it is quite natural and plausible that
their Cartesian product is nonempty. There should be "tuples" with an
element of each set in each slot; a basic idea, "if objects are there,
we can collect them in a set".

@Is its use vital to prove any vital theorems? IÕm open to a discussion what
@is vital. I suggest a theorem is not made vital by: (1) the fact that
@you learned it in second-year university; (2) itÕs a really cool result,
@one of your personal favorites, and youd like it to be true; or (3) itÕs
@a central theorem of hyper-regular-Banach algebras, and the theory of
@hyper-regular-Banach algebras couldnÕt exist without it. I suggest a
@theorem would be vital if itÕs used by the sciences. Of course, it loses
@vitalness if for the same effect the sciences could use another theorem
@which does not need AC.

No problem with 1,2,3; eh, I was goofing off in my second year.
But "used by the sciences" is not a criterion either. For one thing, the
sciences have often found what they needed, already developed in Mathema-
tics.

@I don't accept arguments by analogy, like "It works with finite sets." If
@that type of argument worked, we would still be saying that all infinite
@series are convergent.

Oh... okay, don't make any arguments by analogy then!

Like many other things, analogy can be used thoughtfully, or
thoughtlessly.

@>And do they apply to
@>just AC_w too? (Choice for countably many sets, E_0, E_1, ...).
@>
@
@Well again, I think you have it backwards. Can you provide a justification
@for AC_w?

Actually, _you_ can. At times you have considered, I imagine,
sequences x_n -> a... argued that the union of countably many countable
sets is countable... selected a sequence of sets... Chances are at least
some of these involved AC_w or its consequences. Most people doing math
have; it was totally natural and obvious to them. They didn't see it as
application of some additional principle.

So people have voted with their feet, validating AC_w.

All right, time to hear the other side.

Ilias

[Fred Galvin]

Brian M. Scott <sc...@math.csuohio.edu> wrote:

>On 9 Sep 1997 04:41:10 GMT, mat...@math.canterbury.ac.nz (Bill
>Taylor) wrote:

> >...we'd have a bunch of black push-me-pull-yous

> >whose ends can't be distinguished.

I think that word is spelled "pushmi-pullyus"; anyway, it's definitely not
"push-me-pull-yous".

>I believe that you're looking at a different theorem. As I understand
>this one, it says that if f : 2 x a --> 2 x b is a bijection, then
>there is a bijection between a and b.

Right. The most general version says that, given an INjection
f : n x A --> n x B, where n is a positive integer, you can define an
INjection g : A --> B. I'm not sure who first put the theorem in this
more general form, with inequalities instead of equalities; Bernstein or
Tarski or someone in between. (Exercise: prove without choice that, if 2a
is less than or equal to 2b, then a is less than or equal to b.)

[Andrew Boucher]

In article <5usfj2$4...@gap.cco.caltech.edu>, ika...@alumni.caltech.edu
(Ilias Kastanas) wrote:

>
>In article <aboucher-060...@cyber133-3.paris.imaginet.fr>,

>Andrew Boucher <abou...@imagiNet.fr> wrote:
>>In article <5uih0j$bme$1...@cantuc.canterbury.ac.nz>,
>>mat...@math.canterbury.ac.nz (Bill Taylor) wrote:
>>

>>>All that's wrong with it (Axiom of Choice), is that it's false. :) And

not
>>even false in set

>>>theory, as the "standard model" V is so non-"real" it doesn't matter much

>>>*what* you consistently assume about it. It's just that it's false in

>>>analysis, and importantly so when used to prove existential results.

>>>(As for universal results - who cares? What matter if every vector space

>>>has a basis, when any particular one you ever work with is certain to

>>>have one anyway...)
>
>>Which approach is preferable:

>> 1) State a theorem generalizing over all vector spaces and in the

first

>>line of proof assert by AC the existence of a basis; or

>> 2) State the theorem generalizing over all vector spaces with

bases and
>>not use AC?
>>
>>Surely 2 is better.
>
>

> I hope you were careful in doing that proof, e.g. in using
dimension.

>Without AC, there can be a vector space that has two bases of different

>cardinalities.

>
>
>>Surely 2 is better. As you say any particular one we work with has a
basis

>>in any case, so the extra generality that 1 brings is not useful--and it
>
>
>

> I've mentioned this before: is it "evident" that R has a basis over
Q?
>

I myself didnÕt mention "evident"--perhaps I missed a message. What

troubles me is a theorem which generalizes over all vector spaces and which

really needs to generalize over all vector spaces with bases. This is done

by invoking AC at the beginning of the proof. Since I find AC questionable,

I donÕt like this.

>
>>comes at the price of assuming an axiom which is questionable.
>
>

> Maybe it's worth discussing on what grounds might AC be

questionable.

>Not having been here too long, I don't recall any such debate. (I hope we

>won't be disturbing threads about a trillion Bible codes forming ley lines

>on the Great Pyramid in 2001).
>

> So then, are there cogent objections to AC?

ArenÕt we getting this backwards? ArenÕt you the one who is assuming

something? So shouldnÕt you be the one to provide reasons why I should

accept it? Otherwise, I should be able to invoke OccamÕs Razor.

Is it intuitively obvious?

Is its use vital to prove any vital theorems? IÕm open to a discussion what

is Òvital.Ó I suggest a theorem is not made vital by: (1) the fact that

you learned it in second-year university; (2) itÕs a really cool result,

one of your personal favorites, and youÕd like it to be true; or (3) itÕs

a central theorem of hyper-regular-Banach algebras, and the theory of

hyper-regular-Banach algebras couldnÕt exist without it. I suggest a

theorem would be vital if itÕs used by the sciences. Of course, it loses

vitalness if for the same effect the sciences could use another theorem

which does not need AC.

I don't accept arguments by analogy, like "It works with finite sets." If

that type of argument worked, we would still be saying that all infinite

series are convergent.

>And do they apply to

>just AC_w too? (Choice for countably many sets, E_0, E_1, ...).
>

Well again, I think you have it backwards. Can you provide a justification
for AC_w?

>
>
> Ilias

[Fred Galvin]

On 9 Sep 1997, Bill Taylor wrote:

> Well, there's a subtle point here, isn't there? This gives set 2a as a+a.

> i.e. 2 lots of a. I guess this is what 2a means? But it is not necessarily

> equal to a2, is it, in the absence of choice? That (a2) would be a lots of 2,
> i.e. "a" lots of unordered pairs, which couldn't automatically be classified
> as heads and tails. (Who brought coin-tossing into this!) That is, instead
> of a bunch of a black horses, we'd have a bunch of black push-me-pull-yous
> whose ends can't be distinguished. Then I think the proof breaks down.

By definition, ab = |AxB| where |A| = a and |B| = b. No choice is needed
to show that 2a = a2 = a+a for any cardinal a. (Remember, we're talking
*cardinal* arithmetic here, not ordinals.) And you're right, I'm not
saying anything about the union of a collection of unordered pairs; as you
put it, that's a horse of a different color.

> If I'm right, that means that: whereas 2a = 2b ==> a = b,
> it is not so that a2 = b2 ==> a = b,
> Have I got that right, anyone? Interesting.

Well, you're not using the standard definition of ab (I recommend
Sierpinski's _Cardinal and Ordinal Numbers_, very readable, that's where I
learned about this stuff); and, as Ilias pointed out, what you mean by a2
(the cardinality of the union of a 2-element sets) is not well-defined
without choice. Still, one can ask (as I guess you did) whether, in ZF,
the existence of a bijection between the unions of two families of
disjoint 2-element sets implies the existence of a bijection between the
families. I guess not, but I'll wait to see what the real set-theorists
say about that.

> Looking at Fred's tops-&-tails proof, it seems that "all" it boils down to is
> noticing that 2a == 2b means there is an INJECTION of a into b, (just take
> take half of the LHS and compress the RHS in the obvious way). And by symmetry
> there is also an injection of b into a.
> So the choice-free Schroder-Bernstein jumps this hurdle quite easily. There
> was indeed a strong farmyard smell of SB (BS?) about Fred's nice proof anyway.

You lost me here. *If* the construction I described only gave injections,
then Schroder-Bernstein would come to the rescue; but in fact it already
gives you the bijection. If you're saying there is some more obvious way
to define the mapping from a to be if you only need it to be injective, I
don't see that.

It's not my proof, by the way; only the equine imagery (which I'm starting
to regret) is mine. You can give similar proofs using e.g. people or (as
Ilias pointed out) knives and forks.

[##ANDREW## ##BOUCHER##]

In article <5v4n73$hl1$1...@nuke.csu.net> ilias kastanas 08-14-90,

>@I myself didnŐt mention "evident"--perhaps I missed a message. What

>@troubles me is a theorem which generalizes over all vector spaces and which
>@really needs to generalize over all vector spaces with bases. This is done
>@by invoking AC at the beginning of the proof. Since I find AC questionable,

>@I donŐt like this.

>
>
>
> You mentioned that "any particular space we work with has a basis
>in any case". So I asked about R over Q. Is it one of the "vector
>spaces with bases"? (And if yes, does it have dimension? ... c.f. above)
>

Well, usually I do not work with R over Q... If you would allow me to
restate my point, either you can prove a particular vector space V has a
basis (without AC!), in which case the theorem łEvery vector space with a
basis...˛ applies, or you cannot prove it, in which case I am not
convinced by an application of the theorem łEvery vector space...˛ which
begins by appealing to AC.

>
>@>>comes at the price of assuming an axiom which is questionable.
>@>
>@>
>@> Maybe it's worth discussing on what grounds might AC be
>@questionable.
>@>Not having been here too long, I don't recall any such debate. (I hope we
>@>won't be disturbing threads about a trillion Bible codes forming ley lines
>@>on the Great Pyramid in 2001).
>@>
>@> So then, are there cogent objections to AC?
>

>@ArenŐt we getting this backwards? ArenŐt you the one who is assuming
>@something? So shouldnŐt you be the one to provide reasons why I should
>@accept it? Otherwise, I should be able to invoke OccamŐs Razor.

>
>
>
> Uh, if you can provide reasons why I should accept Occam's Razor...
>

General principle: Suppose S is a set of propositions properly contained
in T. Suppose also that both S and T are adequate to prove the theorems
which are of interest. Then S is preferable to T.

For example, suppose I want to do number theory. I have certain theorems
I want to prove in mind. For the sake of argument, the Peano Axioms
prove all of these. Then the Peano Axioms including the Godel statement
(the one which asserts its unprovability by the Peano axioms) also prove
these theorems. Nonetheless, the Peano Axioms are preferable to the
Peano Axioms plus the Godel statement.

One reason is that one of the major purposes of the axiomatic method is,
as Frege says, łto afford us insight into the dependence of truths upon
one another.˛ If you assume an axiom which we do not need, you are
confusing matters.

> No kidding. It's not forwards or backwards; I don't _have_ to
>defend it, and you don't _have_ to attack it. It's all voluntary.
>

Yeah, and we do not have to do any mathematics, either. No one is going
to force anyone, and we can all live in peace and happiness. But I
prefer to assume axioms in which I have some sort of confidence; and that
means, I will not accept (or at least try not to accept!) an axiom unless
I have an acceptable reason for doing so.

>
>
>@Is it intuitively obvious?
>
>
>
> Yes. If sets are nonempty, it is quite natural and plausible that

I get it. łNatural and plausible˛ is the same as łintuitively obvious˛.
So I think you have answered my question as łno˛. After all, you lowered
the bar only because you knew you could not jump the higher one.

>their Cartesian product is nonempty.

Gee whiz, there is the analogy to the finite case I said I would not
accept. Why should the infinite case work like the finite case? There
are numerous instances where the infinite is different than the finite.

>There should be "tuples" with an
>element of each set in each slot; a basic idea, "if objects are there,
>we can collect them in a set".
>

For one thing collecting in tuples is different than collecting in sets,
no?

>
>@Is its use vital to prove any vital theorems? IŐm open to a discussion what

>@is vital. I suggest a theorem is not made vital by: (1) the fact that

>@you learned it in second-year university; (2) itŐs a really cool result,
>@one of your personal favorites, and youd like it to be true; or (3) itŐs

>@a central theorem of hyper-regular-Banach algebras, and the theory of

>@hyper-regular-Banach algebras couldnŐt exist without it. I suggest a
>@theorem would be vital if itŐs used by the sciences. Of course, it loses

Excuse me? This seems to lack a historical perspective. People łwere
voting with their feet˛ when they thought that all infinite series
converged, or all continuous functions were differentiable almost
everywhere, or... Mathematicians can be lemmings just like the rest of
the population.

[David Eppstein]

In <5v7ra3$enj$1...@nuke.csu.net> ika...@sol.uucp (ilias kastanas 08-14-90) writes:
> But what if you are looking for solutions of f(x+y) = f(x) + f(y)
> other than f(x) = cx ? Assuming AC you get a basis for R over Q and find
> lots of them. Assuming "all sets of reals are Lebesgue-measurable" implies
> none exist. So if you work without AC you cannot resolve the question. Or
> take topology; Tychonoff's theorem (product of compact spaces is compact) is
> equivalent to AC.

So, since the Dehn invariants used to solve Hilbert's third problem do
(as far as I remember) involve solutions to equations resembling that
one, do those who believe in the falsity of choice also believe that
regular tetrahedra can be cut up into finitely many polyhedra and
rearranged to form cubes?

(Is it a rarity that AC is used in a nonexistence rather than existence
proof? I'm sure what they really believe is that the problem is still
open, if indeed it is open without some form of choice.)
--
David Eppstein UC Irvine Dept. of Information & Computer Science
epps...@ics.uci.edu http://www.ics.uci.edu/~eppstein/

[ilias kastanas 08-14-90]

In article <5v73il$f8a$1...@belzebul.imaginet.fr>,
##ANDREW## ##BOUCHER## <abou...@imaginet.fr> wrote:
@>@In article <5usfj2$4...@gap.cco.caltech.edu>, ika...@alumni.caltech.edu
@>@(Ilias Kastanas) wrote:
...

@>@> I've mentioned this before: is it "evident" that R has a basis over Q?
@>@

@>@I myself didnÕt mention "evident"--perhaps I missed a message. What
@>@troubles me is a theorem which generalizes over all vector spaces and which
@>@really needs to generalize over all vector spaces with bases. This is done
@>@by invoking AC at the beginning of the proof. Since I find AC questionable,
@>@I donÕt like this.
@>
@>
@> You mentioned that "any particular space we work with has a basis
@>in any case". So I asked about R over Q. Is it one of the "vector
@>spaces with bases"? (And if yes, does it have dimension? ... c.f. above)

@Well, usually I do not work with R over Q... If you would allow me to
@restate my point, either you can prove a particular vector space V has a
@basis (without AC!), in which case the theorem ³Every vector space with a
@basis...² applies, or you cannot prove it, in which case I am not
@convinced by an application of the theorem ³Every vector space...² which
@begins by appealing to AC.

If a proof says "Using AC, let B be a basis for V", and you intend
to apply the theorem to V = R^n over R, it's pretty clear you don't need AC.
Maybe all your work is in areas like that. E.g. for arithmetical statements
provable in standard mathematics (ZF + AC), AC is irrelevant; you might as
well work in ZF alone.

But what if you are looking for solutions of f(x+y) = f(x) + f(y)
other than f(x) = cx ? Assuming AC you get a basis for R over Q and find
lots of them. Assuming "all sets of reals are Lebesgue-measurable" implies
none exist. So if you work without AC you cannot resolve the question. Or
take topology; Tychonoff's theorem (product of compact spaces is compact) is
equivalent to AC.

Since you don't believe AC, you probably stay clear of such areas.
Fine; that's your choice (!).

@>@>>comes at the price of assuming an axiom which is questionable.
@>@>
@>@>
@>@> Maybe it's worth discussing on what grounds might AC be questionable.
@>@>Not having been here too long, I don't recall any such debate. (I hope we
@>@>won't be disturbing threads about a trillion Bible codes forming ley lines
@>@>on the Great Pyramid in 2001).

@>@>
@>@> So then, are there cogent objections to AC?
@>

@>@ArenÕt we getting this backwards? ArenÕt you the one who is assuming
@>@something? So shouldnÕt you be the one to provide reasons why I should
@>@accept it? Otherwise, I should be able to invoke OccamÕs Razor.
@>
@>
@> Uh, if you can provide reasons why I should accept Occam's Razor...
@>
@
@
@General principle: Suppose S is a set of propositions properly contained
@in T. Suppose also that both S and T are adequate to prove the theorems
@which are of interest. Then S is preferable to T.
@
@For example, suppose I want to do number theory. I have certain theorems
@I want to prove in mind. For the sake of argument, the Peano Axioms
@prove all of these. Then the Peano Axioms including the Godel statement
@(the one which asserts its unprovability by the Peano axioms) also prove
@these theorems. Nonetheless, the Peano Axioms are preferable to the
@Peano Axioms plus the Godel statement.
@
@One reason is that one of the major purposes of the axiomatic method is,
@as Frege says, ³to afford us insight into the dependence of truths upon
@one another.² If you assume an axiom which we do not need, you are
@confusing matters.

This was a joke on "provide reasons"...

...
@>@Is it intuitively obvious?
@>
@>
@> Yes. If sets are nonempty, it is quite natural and plausible that
@
@I get it. Natural and plausible is the same as intuitively obvious.
@So I think you have answered my question as no. After all, you lowered
@the bar only because you knew you could not jump the higher one.

Eh? Yes, "natural and plausible" does seem to mean "intuitively
obvious"; and in any case, that is what _I_ meant by it.

@>their Cartesian product is nonempty.
@
@
@Gee whiz, there is the analogy to the finite case I said I would not
@accept. Why should the infinite case work like the finite case? There
@are numerous instances where the infinite is different than the finite.

Now you are imagining things. What analogy? I'm talking about
any family, finite or infinite, of whatever sets, finite or infinite. The
family may be finite, countable (AC_w), or arbitrary (AC); I'm making the
same, intuitive claim for all.

@>There should be "tuples" with an
@>element of each set in each slot; a basic idea, "if objects are there,
@>we can collect them in a set".
@>
@
@For one thing collecting in tuples is different than collecting in sets, no?

Same result. Let me state this in detail:

The elements of a set F are nonempty sets, and disjoint (for any A
in F and B in F, B != A, the intersection of A and B is empty). A set Z is
a "choice set for F" if Z contains exactly one element of A, of B ... i.e.
for any A in F, Z intersect A is a singleton, {a}.

"For any F as above, there is a choice set" is equivalent to AC.

In other words, collecting into a set is sufficient to produce AC.

So the claim is, if for each A there is an a in A, we can collect
such a's into a set Z... by the intuition that objects can be collected
into a set. If you don't find this intuitive, you might want to say why.

...
@>@>And do they apply to
@>@>just AC_w too? (Choice for countably many sets, E_0, E_1, ...).
@>@>
@>@

@>@Well again, I think you have it backwards. Can you provide a justification

@>@for AC_w?
@>
@>
@> Actually, _you_ can. At times you have considered, I imagine,
@>sequences x_n -> a... argued that the union of countably many countable
@>sets is countable... selected a sequence of sets... Chances are at least
@>some of these involved AC_w or its consequences. Most people doing math
@>have; it was totally natural and obvious to them. They didn't see it as
@>application of some additional principle.
@>
@> So people have voted with their feet, validating AC_w.
@>
@
@Excuse me? This seems to lack a historical perspective. People ³were
@voting with their feet² when they thought that all infinite series
@converged, or all continuous functions were differentiable almost
@everywhere, or... Mathematicians can be lemmings just like the rest of
@the population.

_Who_ thought that all infinite series converge? Euler dabbled
with assigning a possible "limit" to 1-1+1-1... and others; but so what?
People suspected differentiability a.e., but did anybody claim a proof?

Besides, this is not what I'm talking about. Let me be specific.
Suppose A is a set of reals, x is not in A, and x is in the closure of
A (every open interval around x meets A). Then there is a sequence a_n
of points in A such that a_n -> x. Or: suppose (0,1) is partitioned
into sets A_1, A_2, ...; then there is a sequence a_n with a_k in A_k.

Do you consider such deductions valid? Most people do, and use
them. That endorses AC_w.

Ilias

[Bill Taylor]

ika...@sol.uucp (ilias kastanas 08-14-90) writes:

|> > Is [AC] intuitively obvious?

|>
|> Yes. If sets are nonempty, it is quite natural and plausible that
|> their Cartesian product is nonempty. There should be "tuples" with an
|> element of each set in each slot; a basic idea,

Ugh! I hate this, ever since it dawned on me how I'd been brainwashed by it.
And lots of other folk too.

Yes, the idea of the silliness of non-existence of an infinity-tuple is a
stock one, to impresss on students the intuitiveness of choice.

But it's a bit of a have. All this does is make intuitively reasonable
the idea of ACw. (And even then examples are likely to be shown of sets
with plenty of internal structure, which makes it easy.) But the
reasonableness of AC is much beyond the reasonableness of ACw;
(which is suspect anyway, in general, (as opposed to with the nice
sets one gets in analysis)).

My guess is that most students who swallow ACw and its infinite-tuples,
would be substantially more likely to gag, faced with the possibility of
a 0-1 function on the unit interval which had "about half of each" in
each sub-interval.

Still, intuition is a very subjective thing, all said and done.

Live and let live, by all means. Just say when it's been used, is all;
and be prepared to look at distinctions in its absence.

-------------------------------------------------------------------------------
Bill Taylor W.Ta...@math.canterbury.ac.nz
-------------------------------------------------------------------------------

I don't give in to peer pressure unless my friends tell me to.
-------------------------------------------------------------------------------

[Ilias Kastanas]

In article <5v7vh2$2...@euclid.ics.uci.edu>,
David Eppstein <epps...@euclid.ics.uci.edu> wrote:

>In <5v7ra3$enj$1...@nuke.csu.net> ika...@sol.uucp (ilias kastanas 08-14-90) writes:
>> But what if you are looking for solutions of f(x+y) = f(x) + f(y)
>> other than f(x) = cx ? Assuming AC you get a basis for R over Q and find
>> lots of them. Assuming "all sets of reals are Lebesgue-measurable" implies
>> none exist. So if you work without AC you cannot resolve the question. Or
>> take topology; Tychonoff's theorem (product of compact spaces is compact) is
>> equivalent to AC.
>

>So, since the Dehn invariants used to solve Hilbert's third problem do
>(as far as I remember) involve solutions to equations resembling that
>one, do those who believe in the falsity of choice also believe that
>regular tetrahedra can be cut up into finitely many polyhedra and
>rearranged to form cubes?
>
>(Is it a rarity that AC is used in a nonexistence rather than existence
>proof? I'm sure what they really believe is that the problem is still
>open, if indeed it is open without some form of choice.)
>--
>David Eppstein UC Irvine Dept. of Information & Computer Science
>epps...@ics.uci.edu http://www.ics.uci.edu/~eppstein/

Well, I would hate to think of anybody busy cutting up tetrahedra!

You are right, in general they would consider a consequence of AC
as open. But Tychonoff's theorem they would consider false.

Existence and nonexistence can be interchangeable, at the price of
some artificiality. Existence of a non-measurable set / non-existence of
a function splitting each set into a Borel set and a set of measure zero.

I don't recall whether choice was used in the solution of Hilbert's
third problem. But the statement "there exists a tetrahedron such that
for any finite decomposition, and any finite decomposition of an equal-volume
cube, there is no matching of congruent pieces" admits elimination anyway;
due to its logical form, if it has a proof in ZF + AC it also has one in ZF.

Ilias

[John Rickard]

David Eppstein (epps...@euclid.ics.uci.edu) wrote:
: So, since the Dehn invariants used to solve Hilbert's third problem do

: (as far as I remember) involve solutions to equations resembling that
: one, do those who believe in the falsity of choice also believe that
: regular tetrahedra can be cut up into finitely many polyhedra and
: rearranged to form cubes?

No; the Dehn invariant takes values in R # (R/pi), where "#" denotes
tensor product over Z; this does not need AC to work. When applying
it to prove that (say) a cube and regular tetrahedron are not
equidecomposable, it may be simplest to use AC to show that there is a
map R # (R/pi) -> R that takes different values on the two values of
the Dehn invariant; however, one can easily avoid AC by restricting
one's attention to finite-dimensional subspaces U and V of R (regarded
as a vector space over Q) such that the two values lie in U # (V/pi),
and creating a map U # (V/pi) -> R instead.

: (Is it a rarity that AC is used in a nonexistence rather than existence
: proof?

A square cannot be divided into an odd number of triangles all of
which have the same area; the proof (or at least a proof) of this uses
AC, but in a way that can easily be eliminated in much the same way as
above.

--
John Rickard

[Matthew P Wiener]

In article <5v7vh2$2...@euclid.ics.uci.edu>, eppstein@euclid (David Eppstein) writes:
>So, since the Dehn invariants used to solve Hilbert's third problem do
>(as far as I remember) involve solutions to equations resembling that
>one, do those who believe in the falsity of choice also believe that
>regular tetrahedra can be cut up into finitely many polyhedra and
>rearranged to form cubes?

>(Is it a rarity that AC is used in a nonexistence rather than existence

>proof? I'm sure what they really believe is that the problem is still
>open, if indeed it is open without some form of choice.)

On this note, one should recall the original Banach-Tarski paradox paper
used AC for _two_ results: the infamous 3D paradox, of course, but also
2D is _not_ similarly paradoxical.

It was almost 20 years before a ZF proof of their 2D nonparadox result
was found.

[A Boucher]

In article <5v7ra3$enj$1...@nuke.csu.net> ilias kastanas 08-14-90,
ika...@sol.uucp writes:

> This was a joke on "provide reasons"...

I have no sense of humor. I am also confused. Do you accept Occam's
Razor or not?

> Eh? Yes, "natural and plausible" does seem to mean "intuitively
>obvious"; and in any case, that is what _I_ meant by it.
>

Okay, you can mean whatever you want. Just donąt expect other people to
understand it the first time around. For curiosityąs sake, do you find
Euclidąs Parallel Line Postulate also łintuitively obvious˛? It seems
fairly łnatural and plausible˛ to me!

>
>@>There should be "tuples" with an
>@>element of each set in each slot; a basic idea, "if objects are there,
>@>we can collect them in a set".
>@>
>@
>@For one thing collecting in tuples is different than collecting in sets, no?
>
>
> Same result. Let me state this in detail:
>
> The elements of a set F are nonempty sets, and disjoint (for any A
>in F and B in F, B != A, the intersection of A and B is empty). A set Z is
>a "choice set for F" if Z contains exactly one element of A, of B ... i.e.
>for any A in F, Z intersect A is a singleton, {a}.
>
> "For any F as above, there is a choice set" is equivalent to AC.
>
> In other words, collecting into a set is sufficient to produce AC.
>
>
> So the claim is, if for each A there is an a in A, we can collect
>such a's into a set Z... by the intuition that objects can be collected
>into a set. If you don't find this intuitive, you might want to say why.
>

I think youąre missing the point. But then it was a small point to
begin with.... ! (1) If you have some objects, use a lasso to łcollect˛
them into a set--fine if you like it. But to my taste your example means
you have to pick out the objects before you can collect them--it is the
picking out, not the collecting, to which I objected. (2) As you know,
a tuple is ordered while a set is not. Your comment switched from
talking about Cartesian products and (ordered) tuples to talking about
collecting objects into an (unordered) set. Sure you can restate AC in
terms of sets rather than tuples, but that wasnąt your point--it was that
Cartesian products and tuples provide a justification for AC. So
logically you should be saying how obvious it is to collect objects into
infinite tuples: one in this slot, one in this, one in this, etc. But
Iąm afraid the obviousness of this in the infinite case escapes me. But
perhaps because it is works in the finite case, you want to argue by
analogy?

> _Who_ thought that all infinite series converge? Euler dabbled
>with assigning a possible "limit" to 1-1+1-1... and others; but so what?

You were right I was thinking about Euler, and right (now that I've
checked!) that that wasn't his mistake. Actually, it was that "he
assumed that all methods of summation were regular" (p. 70, The
Development of the Foundations of Mathematical Analysis from Euler to
Riemann by I Grattan-Guiness). Still wrong, though.

As to the "so what", I'm asking what justifies AC. You answer lots of
mathematicians use it. Well, I'm not an expert in the history of
mathematics, but if Euler made a mistake, probably lots of mathematicians
at the same time made the same mistake. So lots of mathematicians can be
wrong about something. If about infinite series, why not about AC?

>People suspected differentiability a.e., but did anybody claim a proof?

Not that I know of, but I imagine someone--although perhaps not a
well-known mathematician--did.

> Do you consider such deductions valid?

If a deduction uses AC_w, I do not accept it.
>
> Ilias

[WJ Bland]

On 11 Sep 1997, A Boucher wrote:

> You were right I was thinking about Euler, and right (now that I've
> checked!) that that wasn't his mistake. Actually, it was that "he
> assumed that all methods of summation were regular" (p. 70, The
> Development of the Foundations of Mathematical Analysis from Euler to
> Riemann by I Grattan-Guiness). Still wrong, though.
>
> As to the "so what", I'm asking what justifies AC. You answer lots of
> mathematicians use it. Well, I'm not an expert in the history of
> mathematics, but if Euler made a mistake, probably lots of mathematicians
> at the same time made the same mistake. So lots of mathematicians can be
> wrong about something. If about infinite series, why not about AC?

You cannot make a "mistake" about using AC, since it's *INDEPENDENT* of
the rest of standard set theory. Euler made a mistake because he assumed
something to be true when it was actually (*provably*) false.
The same mistake cannot be made in assuming AC, because we can never,
ever prove that AC is false.

Use it if you like it, don't if you don't. AC has no definite truth
value. A hard-core Platonist would disagree with this, but then everyone
knows that Platonists are just silly ;-)

Cheers,
Bill.

[Ilias Kastanas]

In article <5v9prr$r09$1...@belzebul.imaginet.fr>,

A Boucher <abou...@imaginet.fr> wrote:
>In article <5v7ra3$enj$1...@nuke.csu.net> ilias kastanas 08-14-90,
>ika...@sol.uucp writes:
>
>> This was a joke on "provide reasons"...
>
>I have no sense of humor. I am also confused. Do you accept Occam's
>Razor or not?

Well, not a very funny joke anyway.

No dispute regarding unnecessary hypotheses.

>> Eh? Yes, "natural and plausible" does seem to mean "intuitively
>>obvious"; and in any case, that is what _I_ meant by it.

>Okay, you can mean whatever you want. Just donąt expect other people to
>understand it the first time around. For curiosityąs sake, do you find
>Euclidąs Parallel Line Postulate also łintuitively obvious˛? It seems
>fairly łnatural and plausible˛ to me!

Okay, but it wasn't exactly "whatever I want", but a matter of
general understanding. And, whatever the qualifying adjectives, a
specific statement followed.

The Postulate? For the Euclidean plane, yes; for the hyperbolic,
no... The point is, it's not the same thing. There are many kinds of
geometries, or measures, or modules; but the operation of forming sets
appears to have something fundamental about it.

As to (2): Tuples vs sets doesn't matter. I don't see "slots"
as adding something; the basic idea is "collecting". Uh, why do you
keep bringing up "analogy"... which you reject anyway? I don't reject
it myself, but in this case I am simply not using any analogy; what I
wrote claims to justify _all_ cases: finite or infinite family, finite
or infinite sets.

As to (1): Lasso, invisible bag... whatever works! Now, your
objection is in "choosing" -- from infinitely many sets; you have no
problem from finitely many. Each set has at least one element, but
how do we actually pick the one we will "collect"?

I don't really see the (small?) point: Suppose we have just
one set; "how" do we pick an element? There is no method, or rule in
general; we just accept that we can single out an element of the set.
"There is" such an element... for any set in F. So how does this sudden-
ly become objectionable if applied to all sets in F? Choice from any
A is independent of choice from any other A'. We can do any one...
and yet we cannot do them all? Why?

>> _Who_ thought that all infinite series converge? Euler dabbled
>>with assigning a possible "limit" to 1-1+1-1... and others; but so what?
>

>You were right I was thinking about Euler, and right (now that I've
>checked!) that that wasn't his mistake. Actually, it was that "he
>assumed that all methods of summation were regular" (p. 70, The
>Development of the Foundations of Mathematical Analysis from Euler to
>Riemann by I Grattan-Guiness). Still wrong, though.
>
>As to the "so what", I'm asking what justifies AC. You answer lots of
>mathematicians use it. Well, I'm not an expert in the history of
>mathematics, but if Euler made a mistake, probably lots of mathematicians
>at the same time made the same mistake. So lots of mathematicians can be
>wrong about something. If about infinite series, why not about AC?

Eh, this is not an argument against AC, it's against everything
in mathematics! "Everybody could be wrong" about anything in particular...

>>People suspected differentiability a.e., but did anybody claim a proof?
>
>Not that I know of, but I imagine someone--although perhaps not a
>well-known mathematician--did.

Perhaps. But what I mentioned were unlike nebulous claims, or mis-
taken views; they were ways of reasoning, accepted and in standard use.

>> Do you consider such deductions valid?
>
>If a deduction uses AC_w, I do not accept it.

OK; (0,1) is partitioned into sets E0, E1, E2, ... ; claim: there is
a sequence x0, x1, x2,... (xk in Ek, for k in N); it needs AC_w, so you do not
accept this claim. Any specific reasons you can mention?

A0, A1, ... are countable sets. Claim: their union (of Ak, k in N)
is countable; it follows from AC_w. Do you reject it? Any reasons?

Same, with each of A0, A1... having two elements. Now what?

E is infinite. Claim: there is a sequence x0, x1, ... of distinct
elements of E. (Follows from AC_w; weaker than it). What about this?

Such reasoning does come up often!

I don't mean "you should accept these!". I only mean what I said:
if you don't, could you explain your thinking?

Ilias

[Ilias Kastanas]

In article <5v81ts$5ci$1...@cantuc.canterbury.ac.nz>,

Bill Taylor <mat...@math.canterbury.ac.nz> wrote:
>ika...@sol.uucp (ilias kastanas 08-14-90) writes:
>

>|> > Is [AC] intuitively obvious?

>|>
>|> Yes. If sets are nonempty, it is quite natural and plausible that
>|> their Cartesian product is nonempty. There should be "tuples" with an
>|> element of each set in each slot; a basic idea,
>

>Ugh! I hate this, ever since it dawned on me how I'd been brainwashed by it.
>And lots of other folk too.
>
>Yes, the idea of the silliness of non-existence of an infinity-tuple is a
>stock one, to impresss on students the intuitiveness of choice.
>
>But it's a bit of a have. All this does is make intuitively reasonable
>the idea of ACw. (And even then examples are likely to be shown of sets
>with plenty of internal structure, which makes it easy.) But the
>reasonableness of AC is much beyond the reasonableness of ACw;
>(which is suspect anyway, in general, (as opposed to with the nice
>sets one gets in analysis)).
>
>My guess is that most students who swallow ACw and its infinite-tuples,
>would be substantially more likely to gag, faced with the possibility of
>a 0-1 function on the unit interval which had "about half of each" in
>each sub-interval.
>
>Still, intuition is a very subjective thing, all said and done.
>
>Live and let live, by all means. Just say when it's been used, is all;
>and be prepared to look at distinctions in its absence.

Well, Bill, you do know I don't hold back from such -- even below
AC_w... And yes, intuition can be subjective; I hope people who find AC
intuitive as well as people who don't might try to articulate their reasons
-- to the extent it's possible.

"If F is a family of nonempty, disjoint sets there is a set C having
exactly one point in common with each set in F".

This formulation of AC does not mention "tuples", should that make
any difference. I suggested it follows from the intuition "if objects are
there, we can collect them into a set", which seems pretty basic if one
works with sets in the first place.

I avoided indexing on purpose; F could be finite, countable, or
whatever else. This is not about formal facts (finite AC is provable from
ZF, AC_w is not, AC is not provable from ZF + AC_w... all well-known) but
fundamental views. You and others have such against AC, and I would like to
hear what they are. There have been discussions in the past; what is it
like in the present?

Debate, dialogue, assertions etc are likely; but I certainly have
no intention (or interest) to "convert" anybody -- in case it isn't obvious!

Ilias

[A Boucher]

In article <5vbvej$f...@gap.cco.caltech.edu> Ilias Kastanas,
ika...@alumni.caltech.edu writes:

>Suppose we have just
>one set; "how" do we pick an element? There is no method, or rule in
>general; we just accept that we can single out an element of the set.
>"There is" such an element... for any set in F. So how does this sudden-
>ly become objectionable if applied to all sets in F? Choice from any
>A is independent of choice from any other A'. We can do any one...
>and yet we cannot do them all? Why?

I'm not sure I fully understand your question because I think you
understand my reply already in advance, but anyway here goes... Suppose
S is the singleton set containing s. If you can prove Ex(x belongs to
S), then you can immediately refer to s without assuming any more axioms.
For instance, if you want to assert B about s, you could write
"For All x(Bx <--> x is in S)." So there is no problem in saying that
you can pick out s from S.

Nor I would object to you saying you can pick out *one* element in F, or
any one element in A, or any one element in B,... It's picking out the
infinite number which I find dubious.

>In article <5v9prr$r09$1...@belzebul.imaginet.fr>,
>A Boucher <abou...@imaginet.fr> wrote:
>>As to the "so what", I'm asking what justifies AC. You answer lots of
>>mathematicians use it. Well, I'm not an expert in the history of
>>mathematics, but if Euler made a mistake, probably lots of mathematicians
>>at the same time made the same mistake. So lots of mathematicians can be
>>wrong about something. If about infinite series, why not about AC?
>
>
>
> Eh, this is not an argument against AC, it's against everything
>in mathematics! "Everybody could be wrong" about anything in particular...

It's not an argument against everything in mathematics, because I would
never justify *any* axiom by appealing to the fact that most
mathematicians use it. I would justify the axiom "0 is a natural number"
by saying it is intuitively obvious. I would justify other axioms (e.g.
whichever one we use to assume the existence of irrational numbers) by
saying they are useful.

> OK; (0,1) is partitioned into sets E0, E1, E2, ... ; claim: there is
>a sequence x0, x1, x2,... (xk in Ek, for k in N); it needs AC_w, so you do not
>accept this claim. Any specific reasons you can mention?
>
> A0, A1, ... are countable sets. Claim: their union (of Ak, k in N)
>is countable; it follows from AC_w. Do you reject it? Any reasons?
>
> Same, with each of A0, A1... having two elements. Now what?
>
> E is infinite. Claim: there is a sequence x0, x1, ... of distinct
>elements of E. (Follows from AC_w; weaker than it). What about this?
>
>
> Such reasoning does come up often!
>
> I don't mean "you should accept these!". I only mean what I said:
>if you don't, could you explain your thinking?
>

I don't accept the result because of course I don't except AC_w! If you
can prove a proposition using axioms I accept, then of course I would
accept it; but if you cannot, then of course I would not. I don't
believe the contrary (i.e. that there is a counterexample) either,
because I can't prove that either. So I'm agnostic about lots of
propositions.
The reason why I don't accept AC_w or AC (to repeat) is: (1) it is not
intuitively obvious; (2) it is not needed to prove any vital
theorems--vital to science or "everyday" life. When I first wrote my
diatribe, I thought you would object to (2), not (1)! That is, I
expected you to say there are all these very important theorems, and you
couldn't conceive of doing analysis without them, and since analysis is
vital, you win. Then I would have asked you to be very specific, and,
well, you spoiled my fun.

>
> Ilias

[Ilias Kastanas]

In article <Oyd*ER...@news.atml.co.uk>, John Rickard <j...@atml.co.uk> wrote:
>David Eppstein (epps...@euclid.ics.uci.edu) wrote:
>: So, since the Dehn invariants used to solve Hilbert's third problem do

>: (as far as I remember) involve solutions to equations resembling that
>: one, do those who believe in the falsity of choice also believe that
>: regular tetrahedra can be cut up into finitely many polyhedra and
>: rearranged to form cubes?
>

>No; the Dehn invariant takes values in R # (R/pi), where "#" denotes
>tensor product over Z; this does not need AC to work. When applying
>it to prove that (say) a cube and regular tetrahedron are not
>equidecomposable, it may be simplest to use AC to show that there is a
>map R # (R/pi) -> R that takes different values on the two values of
>the Dehn invariant; however, one can easily avoid AC by restricting
>one's attention to finite-dimensional subspaces U and V of R (regarded
>as a vector space over Q) such that the two values lie in U # (V/pi),
>and creating a map U # (V/pi) -> R instead.
>

>: (Is it a rarity that AC is used in a nonexistence rather than existence
>: proof?
>

>A square cannot be divided into an odd number of triangles all of
>which have the same area; the proof (or at least a proof) of this uses
>AC, but in a way that can easily be eliminated in much the same way as
>above.

In fact, it is no accident. Both statements say "For any decom-
position, something" and can be written in Pi-1-1 form (Ax R(x), x real,
R arithmetical. x=y, x < y are such, and operations, Boolean combinations
and any amount of integer quantifiers Am, Em are OK. Pi-1-2 would be
Ax Ey R; Sigma-1-2, Ex Ay R; Pi-1-1 is contained in both).

So a general result applies: if phi is Pi-1-4, and it is a theorem
of ZF + AC (+ GCH, too), then phi is a theorem of ZF. It is guaranteed that
there is a proof without AC, whether we bother to find one or not.

In other words, it is a freebie: for such phi, one might as well
assume AC and GCH! Say, AC wellorders R and allows transfinite induction;
CH even makes the wellordering have type aleph-1. That might be convenient.

It doesn't go any higher, though. There is a Sigma-1-4 theorem
of ZF + AC that is unprovable in ZF.

Ilias

[Ilias Kastanas]

In article <Pine.GSO.3.96.970912095621.23316A-100000@granby>,
WJ Bland <pmy...@unix.ccc.nottingham.ac.uk> wrote:

>On 11 Sep 1997, A Boucher wrote:
>
>> As to the "so what", I'm asking what justifies AC. You answer lots of
>> mathematicians use it. Well, I'm not an expert in the history of

Not quite, i.e. for AC. I said Choice is intuitive; but that
intuition doesn't extend to the forms most often used (Zorn, Wellordering,
...). It takes proof to show equivalence. At some point it does become
clear they are all "the same thing"; but I'm not saying it's obvious up front.

I meant AC_w and consequences: they are in universal use, with full
certainty that they are valid and correct ways to reason. Countable unions
of countable (or finite!) sets are countable; an infinite set has a subset
{x_0, x_1,...}; and so on. That's people in all areas, and not involved with
logic or set theory.

>> mathematics, but if Euler made a mistake, probably lots of mathematicians
>> at the same time made the same mistake. So lots of mathematicians can be
>> wrong about something. If about infinite series, why not about AC?
>

>You cannot make a "mistake" about using AC, since it's *INDEPENDENT* of
>the rest of standard set theory. Euler made a mistake because he assumed
>something to be true when it was actually (*provably*) false.
>The same mistake cannot be made in assuming AC, because we can never,
>ever prove that AC is false.
>
>Use it if you like it, don't if you don't. AC has no definite truth
>value. A hard-core Platonist would disagree with this, but then everyone
>knows that Platonists are just silly ;-)

Yeah, they make all kinds of sweeping generalizations!...

Independence is a fact; the issue is, what are today's views
about its plausibility, value, etc... and do they have coherent reasons
behind them.

Ilias

[A Boucher]

In article <5vccb8$gkh$1...@belzebul.imaginet.fr> A Boucher,

abou...@imaginet.fr writes:
>"For All x(Bx <--> x is in S)."

Whoops! That of course should be "For All x(x is in S --> Bx)."

[Bill Taylor]

A Boucher <abou...@imaginet.fr> writes:

|> do you find
|> Euclidąs Parallel Line Postulate also łintuitively obvious˛? It seems
|> fairly łnatural and plausible˛ to me!

TOUCHE' !!

Give that man a cigar! Mind you, its possible implausibility in the
clumsy form Euclid felt obliged (by finitary considerations) to put it in,
contributed greatly to a major revolution in math.

As I said once before...

Euclid's dedication to forbid anything like a completed infinity may have
been philosophically pure, but was definitely mathematically unfortunate, it
seems to me. If he'd been prepared to conceive of a line as being all of
its infinite extension both ways, then he would never have needed to have
his original clumsy, rather repellant, and noticeably non-self-evident wording
for the parallel axiom. He then might well have come up with something like

Playfair's Axiom: there exists one and only one line through a
~~~~~~~~~~~~~~~~ given point parallel to a given line.

I've always thought this was exceptionally "self-evident" (from a pre-modern
viewpoint); indeed, like it's name, it's all very British and "playing fair"
and clear-cut and obvious! I've often thought that if Euclid had been able
to word the axiom this way, there would have been no age-long attempts to
try to derive it from the others, on the grounds of its non-obviousness.

Who knows how math would have developed then? The discovery of non-Euclidean
geometries may never have been such a scandal.

-------------------------------------------------------------------------------
Bill Taylor W.Ta...@math.canterbury.ac.nz
-------------------------------------------------------------------------------

If I have seen less far than other men,
it is by standing in the footprints of giants.
-------------------------------------------------------------------------------

[Fred Galvin]

On 11 Sep 1997, A Boucher wrote:

> Do you accept Occam's Razor or not?

Don't know if he does, but I do: "entities should not be multiplied
unnecessarily." And, since I accept Occam's Razor, I have to accept the
axiom of choice, because the axiom of choice (in its strongest form, a
definable well-ordering of the universe) follows from Goedel's axiom of
constructibility, and the axiom of constructibility is required by Occam's
Razor. If nonconstructible sets are not "unnecessary entities", please
tell me what they are needed for. Real math and science uses, not fancy
highbrow set-theory-for-set-theory's-sake.

[Brian M. Scott]

On 12 Sep 1997 21:31:20 GMT, A Boucher <abou...@imaginet.fr> wrote:

>In article <5vbvej$f...@gap.cco.caltech.edu> Ilias Kastanas,
>ika...@alumni.caltech.edu writes:

>>Suppose we have just
>>one set; "how" do we pick an element? There is no method, or rule in
>>general; we just accept that we can single out an element of the set.
>>"There is" such an element... for any set in F. So how does this sudden-
>>ly become objectionable if applied to all sets in F? Choice from any
>>A is independent of choice from any other A'. We can do any one...
>>and yet we cannot do them all? Why?

>I'm not sure I fully understand your question because I think you

>understand my reply already in advance, but anyway here goes... Suppose
>S is the singleton set containing s. If you can prove Ex(x belongs to
>S), then you can immediately refer to s without assuming any more axioms.
> For instance, if you want to assert B about s, you could write
>"For All x(Bx <--> x is in S)." So there is no problem in saying that
>you can pick out s from S.

If I've understood you correctly, S = {s}, which is certainly a very
special case. Your explanation of why you see no problem in picking
an element from S doesn't obviously extend to finite sets of
cardinality greater than 1.

>Nor I would object to you saying you can pick out *one* element in F, or
>any one element in A, or any one element in B,... It's picking out the
>infinite number which I find dubious.

I don't understand what makes it different. The choices are
independent, so you may consider making them in parallel (i.e.,
simultaneously). Why is making infinitely many choices more dubious
than making A(N, N) choices, where A is the Ackerman function and N is
the number of elementary particles in the known universe?

>>>As to the "so what", I'm asking what justifies AC. You answer lots of
>>>mathematicians use it. Well, I'm not an expert in the history of
>>>mathematics, but if Euler made a mistake, probably lots of mathematicians
>>>at the same time made the same mistake. So lots of mathematicians can be
>>>wrong about something. If about infinite series, why not about AC?

>> Eh, this is not an argument against AC, it's against everything
>>in mathematics! "Everybody could be wrong" about anything in particular...

>It's not an argument against everything in mathematics, because I would

>never justify *any* axiom by appealing to the fact that most
>mathematicians use it. I would justify the axiom "0 is a natural number"
>by saying it is intuitively obvious.

It seems to me to be purely a matter of definition.

>I would justify other axioms (e.g.
>whichever one we use to assume the existence of irrational numbers) by
>saying they are useful.

But you don't accept a justification of AC on the grounds that it's
useful?

> The reason why I don't accept AC_w or AC (to repeat) is: (1) it is not
>intuitively obvious; (2) it is not needed to prove any vital
>theorems--vital to science or "everyday" life. When I first wrote my
>diatribe, I thought you would object to (2), not (1)!

(1) is clearly a matter of opinion: I find AC intuitively obvious. I
don't care about (2): the restriction of mathematics to the part known
(at any given moment) to be applicable to science or everyday life
seems to me wholly artificial.

Brian M. Scott

[Ilias Kastanas]

In article <5vccb8$gkh$1...@belzebul.imaginet.fr>,

A Boucher <abou...@imaginet.fr> wrote:
>In article <5vbvej$f...@gap.cco.caltech.edu> Ilias Kastanas,
>ika...@alumni.caltech.edu writes:
>

>>Suppose we have just
>>one set; "how" do we pick an element? There is no method, or rule in
>>general; we just accept that we can single out an element of the set.
>>"There is" such an element... for any set in F. So how does this sudden-
>>ly become objectionable if applied to all sets in F? Choice from any
>>A is independent of choice from any other A'. We can do any one...
>>and yet we cannot do them all? Why?

>I'm not sure I fully understand your question because I think you

>understand my reply already in advance, but anyway here goes... Suppose
>S is the singleton set containing s. If you can prove Ex(x belongs to
>S), then you can immediately refer to s without assuming any more axioms.
> For instance, if you want to assert B about s, you could write
>"For All x(Bx <--> x is in S)." So there is no problem in saying that
>you can pick out s from S.

Here, it's part of the assumption that each A is nonempty; we
have Ex (x in A).

>Nor I would object to you saying you can pick out *one* element in F, or
>any one element in A, or any one element in B,... It's picking out the
>infinite number which I find dubious.

I understand; I'm trying to find out how this comes to terms with
the fact that at each A, there is an x. That's infinitely many x, yes; why
is that an obstacle to putting them in a set C? E.g. we say "let D be
an infinite subset of [0,1]..." and we go ahead to prove it has a limit
point. Well then, we just put infinitely many points into D. Why isn't
_that_ objectionable?

I'm looking for some coherence. It is acceptable that an element
can be picked out of each A; why would "the set of them" be problematic?
As if all of a sudden we had to run huffing and puffing from one A to
another and build C manually!

>>In article <5v9prr$r09$1...@belzebul.imaginet.fr>,
>>A Boucher <abou...@imaginet.fr> wrote:
>>>As to the "so what", I'm asking what justifies AC. You answer lots of
>>>mathematicians use it. Well, I'm not an expert in the history of
>>>mathematics, but if Euler made a mistake, probably lots of mathematicians
>>>at the same time made the same mistake. So lots of mathematicians can be
>>>wrong about something. If about infinite series, why not about AC?
>>
>>
>> Eh, this is not an argument against AC, it's against everything
>>in mathematics! "Everybody could be wrong" about anything in particular...
>

>It's not an argument against everything in mathematics, because I would
>never justify *any* axiom by appealing to the fact that most
>mathematicians use it. I would justify the axiom "0 is a natural number"

>by saying it is intuitively obvious. I would justify other axioms (e.g.

>whichever one we use to assume the existence of irrational numbers) by
>saying they are useful.

The point is, mathematicians don't just use it, they also find it
intuitively obvious. E.g. the four cases below.

Usefulness. Would you accept something not intuitively obvious
because it's useful? (And in what sense?)

>> OK; (0,1) is partitioned into sets E0, E1, E2, ... ; claim: there is
>>a sequence x0, x1, x2,... (xk in Ek, for k in N); it needs AC_w, so you do not
>>accept this claim. Any specific reasons you can mention?
>>
>> A0, A1, ... are countable sets. Claim: their union (of Ak, k in N)
>>is countable; it follows from AC_w. Do you reject it? Any reasons?
>>
>> Same, with each of A0, A1... having two elements. Now what?
>>
>> E is infinite. Claim: there is a sequence x0, x1, ... of distinct
>>elements of E. (Follows from AC_w; weaker than it). What about this?
>>
>>
>> Such reasoning does come up often!
>>
>> I don't mean "you should accept these!". I only mean what I said:
>>if you don't, could you explain your thinking?

>I don't accept the result because of course I don't except AC_w! If you
>can prove a proposition using axioms I accept, then of course I would
>accept it; but if you cannot, then of course I would not. I don't
>believe the contrary (i.e. that there is a counterexample) either,
>because I can't prove that either. So I'm agnostic about lots of
>propositions.

None of the above are either provable or refutable from "mathematics
without choice"; that has been proven.

> The reason why I don't accept AC_w or AC (to repeat) is: (1) it is not
>intuitively obvious; (2) it is not needed to prove any vital
>theorems--vital to science or "everyday" life. When I first wrote my

>diatribe, I thought you would object to (2), not (1)! That is, I
>expected you to say there are all these very important theorems, and you
>couldn't conceive of doing analysis without them, and since analysis is
>vital, you win. Then I would have asked you to be very specific, and,
>well, you spoiled my fun.

Sorry about that. Let us say, you did use choice when you learned
calculus. E.g. "if for every x_n -> c, f(x_n) -> f(c) then f is conti-
nuous at c". Or "a set of reals E is closed and bounded iff every sequence
of points of E has a convergent subsequence". I don't suppose you'll stop
using these theorems, and their consequences?

Anyway, since you seem to reject that (1) the union of countably many
two-element sets is countable, or (2) any infinite set has a countable subset
x_0, x_1, ... , I asked you to explain, if possible, what makes you see
them as non-obvious or doubtful.

Ilias

[Ilias Kastanas]

In article <5vd6it$aeg$3...@cantuc.canterbury.ac.nz>,

Bill Taylor <mat...@math.canterbury.ac.nz> wrote:
>A Boucher <abou...@imaginet.fr> writes:
>
>|> do you find
>|> Euclidąs Parallel Line Postulate also łintuitively obvious˛? It seems
>|> fairly łnatural and plausible˛ to me!
>
>TOUCHE' !!
>
>Give that man a cigar! Mind you, its possible implausibility in the
>clumsy form Euclid felt obliged (by finitary considerations) to put it in,
>contributed greatly to a major revolution in math.

Eh... okay, so what is it, plausible or implausible?!

Of course that an axiom was needed for that in the first place was
quite non-obvious at the time.

>As I said once before...
>
>Euclid's dedication to forbid anything like a completed infinity may have
>been philosophically pure, but was definitely mathematically unfortunate, it
>seems to me. If he'd been prepared to conceive of a line as being all of
>its infinite extension both ways, then he would never have needed to have
>his original clumsy, rather repellant, and noticeably non-self-evident wording
>for the parallel axiom. He then might well have come up with something like

[heresy snipped]

I wonder about that, Bill. One can see "finitary" leanings in
"there is no greatest prime", rather than "there are infinitely many"...
(And it has been noticed that Euclid never used AC...)

But here? Euclid's 5th postulate did not emerge in a vacuum; the
man gave us propositions 27 and 28, where he says it: the lines will not
meet. He is discussing this in terms of a third line, "falling" on them;
having covered the "2 right angles" case (parallelism), he says in his
Fifth that in the "less than 2 r.a." case the lines meet -- and where.
It is the same figure and setting.

Even on a first reading, Euclid comes through clearly: one line
is parallel, and the others aren't. I note that O. Veblen ("Foundations")
takes 2 lines to get from the 5th to "only one parallel"; for the other dire-
ction though he needs half a page of proof. So maybe, economy of logic?!

>I've always thought this was exceptionally "self-evident" (from a pre-modern
>viewpoint); indeed, like it's name, it's all very British and "playing fair"
>and clear-cut and obvious! I've often thought that if Euclid had been able
>to word the axiom this way, there would have been no age-long attempts to
>try to derive it from the others, on the grounds of its non-obviousness.

I don't know, but aren't you underestimating the drive to do
math... -- in a bathtub, in the middle of the city's fall, on the night
before a duel, manuscripts in a WWII soldier's backpack... Hey, here we
have trillions and Bible codes and 2001 cosmic units of inertia, lasting
forever; people just can't get away from _them_... I doubt Saccheri would
do crosswords.

Hmm... we have weak choice too...

Ilias

[Herman Rubin]

In article <5vd6it$aeg$3...@cantuc.canterbury.ac.nz>,
Bill Taylor <mat...@math.canterbury.ac.nz> wrote:
>A Boucher <abou...@imaginet.fr> writes:

.................

>Euclid's dedication to forbid anything like a completed infinity may have
>been philosophically pure, but was definitely mathematically unfortunate, it
>seems to me. If he'd been prepared to conceive of a line as being all of
>its infinite extension both ways, then he would never have needed to have
>his original clumsy, rather repellant, and noticeably non-self-evident wording
>for the parallel axiom. He then might well have come up with something like

>Playfair's Axiom: there exists one and only one line through a

>~~~~~~~~~~~~~~~~ given point parallel to a given line.

Remember that Euclid wrote well before the idea of variable.

By the post-Renaissance time, when people looked at the question of
whether it was provable, this is essentially what they looked at.

>I've always thought this was exceptionally "self-evident" (from a pre-modern
>viewpoint); indeed, like it's name, it's all very British and "playing fair"
>and clear-cut and obvious! I've often thought that if Euclid had been able
>to word the axiom this way, there would have been no age-long attempts to
>try to derive it from the others, on the grounds of its non-obviousness.

I doubt that it would have made any difference in the development. Those
accustomed to reading Euclid, even in translation, in the later times
did not have any problems with this terminology.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558

[Bill Taylor]

ika...@alumni.caltech.edu (Ilias Kastanas) writes:

|> >and be prepared to look at distinctions in its absence.

|> Well, Bill, you do know I don't hold back from such

Indeed I do! You're a good guy though, Ilias, pity there aren't more of you.

|> You and others have such against AC, and I would like to
|> hear what they are. There have been discussions in the past; what is it
|> like in the present?

Well, my motivation is simple enough. It is merely that the only sets
which "really exist" are definable ones; (and quite possibly not even all
of those! But that's a whole nother thread.)

It seems this was Lebesgue's (& Borel's) idea also; but they lived back before
such concepts could be neatly codified, (as e.g. in my w_1^CK post, though
others have done better).

So all those lovely choice-necessitous things - Hamel bases, nonmeasurable
sets, Banach-T partitions, unprincipled infiltrators, etc etc, which ARE
lovely things, no question, simply "aren't there" (in ontological space!)
So results about them, and many results using them, are more like "what it
would be like if these (impossible) things *did* exist". A kind of fantasy
math. What was Weemba's term for it...? Oh yes, "SURREAL math". Nice.

Not that I've anything against it at all. It's all legitimate math, math
doesn't *have* to be fully "real", as we know. I'm all for Hilbert and
the pure math toast! But still, it's irritating to see surreal math
passed off as real math, as is automatically assumed in today's texts. Well,
perhaps not quite automatically... was it Kreisel who said, "AC is unique
in its ability to trouble the conscience of the ordinary mathematician!"

AC_w being used in 19th-century style math is not nearly so heinous; most
such cases (of existence proofs), can easily be made constructive for any
particular example. It's very hard to drum up *really* nonconstructive
things on the real line; the early intuitionists notwithstanding. But no
doubt it could be done, with extraordinary effort; it just won't happen in
bridge-building math. And anyway, there's always the conservative extension
meta-theorem to fall back on!

So.
That's about it. It could be said to be a case of ontological minimality,
to put a pompous label on it, or Occam's razor, to give it a patina of
scholastic respectability. But it just amounts to noticing that non-definable
things "just don't exist", however consistent they may be.

--------

On this general topic, there are some remarkably tricky things proved with AC
in ordinary R^3. Things like, "it is possible to partition 3-D space into
lines all with different slopes"; or "circles all with nonparallel planes";
or other things like so. Alas, my colleague who told me about such matters
many years ago can't recall them now!

Can anyone else help?

-------------------------------------------------------------------------------
Bill Taylor W.Ta...@math.canterbury.ac.nz
-------------------------------------------------------------------------------

There are no objects such that there are no such objects.
-------------------------------------------------------------------------------

[Daryl McCullough]

mat...@math.canterbury.ac.nz (Bill Taylor) says...

>Well, my motivation [for rejecting AC?] is simple enough.

>It is merely that the only sets which "really exist" are
>definable ones; (and quite possibly not even all
>of those! But that's a whole nother thread.)

I have a few comments about that. First: If the only sets
that really exist are the definable ones, then the axiom
of choice *holds*! You can well-order any collection of
sets by well-ordering their definitions. Right.

Second comment: I don't personally see how your intuition
that the only sets that exist are the definable ones
squares with my intuition of positive real numbers as
lengths. Given a physical object, such as a tree or a
book, if you measure its length, I don't see any reason
whatsoever (a priori) to think that the result should
be a definable real. Of course, this intuition of lengths
as being arbitrary breaks down because of atomic theory;
there is really a limit to the meaningful precision in
the length of a physical object. However, I don't see
any reason that this must be the case. I definitely
don't want to skew mathematics so that our current
theory of physics dictates what is mathematically
possible. (Math must be rich enough to handle physics,
but it doesn't have to be a precise fit, in my opinion.)

Third point: I don't understand quite the difference
between your qualms about undefinable sets and the
intuitionists qualms about unconstructable sets. Could
you explain the difference?

>So all those lovely choice-necessitous things - Hamel bases, nonmeasurable
>sets, Banach-T partitions, unprincipled infiltrators, etc etc, which ARE
>lovely things, no question, simply "aren't there" (in ontological space!)

I don't really understand the ontological part of the argument at all.
What does existence mean for abstractions like sets? You seem to say
that things exist if they have names. But, if you just throw in a new
choice operator into set theory (that is, a function symbol F with the
axiom that all x, (x nonempty implies F(x) is an element of x)) then
voila! You have names for all the choice sets. The choice function on
any set S is definable as the graph of F restricted to S.

Daryl McCullough
CoGenTex, Inc.
Ithaca, NY

[Brian M. Scott]

On 15 Sep 1997 19:59:52 GMT, A Boucher <abou...@imaginet.fr> wrote:

>In article <5vdd3s$j...@gap.cco.caltech.edu> Ilias Kastanas,
>ika...@alumni.caltech.edu writes:

>> I'm looking for some coherence. It is acceptable that an element
>>can be picked out of each A; why would "the set of them" be problematic?

>Because you must pick out an infinite number!

You say this as if picking out an infinite number were obviously
problematic. I take it that it is for you, but I still don't see why
it should be (psychologically) different from picking out a very large
finite number.

=====

>> Usefulness. Would you accept something not intuitively obvious
>>because it's useful? (And in what sense?)

>I gave an example: the assumption of the existence of the irrational
>numbers. Okay, it's not the level of sets, but consider axioms for real
>analysis on the level of Apostol (p. 2-9). At some point you have to
>make an assumption which will imply that irrational numbers exist. I
>accept this, even though I do not think it is "intuitively obvious",
>because the existence of irrationals seems to me a "vital" assumption, in
>that it permeates the physicist's model of reality. AC does not cut it
>for me in the same way, although I am open to being corrected. That is,
>if mathematicians stopped using AC and had to add a few more assumptions
>in their propositions (assumptions obeyed in any case by functions used
>by physics, science, and everyday life), I think no one would notice,
>except the mathematician.

How does this make AC less 'vital'?

>> Anyway, since you seem to reject that (1) the union of countably many
>>two-element sets is countable, or (2) any infinite set has a countable subset
>>x_0, x_1, ... , I asked you to explain, if possible, what makes you see
>>them as non-obvious or doubtful.

>I'm really not sure how to reply. Sure, you can whack me over the head
>if I were to claim, "Red is not a color" isn't obvious. But surely I'm
>allowed to claim (1) and (2) are not obvious. Because really and truly,
>they are not obvious to me. And really and truly, I would want a proof
>of these propositions before accepting them. Do you find them so obvious
>that you just accept them without proof!?

(1) certainly seems intuitively obvious. Intuition says that I can go
through and count the union by twos: 2, 4, 6, .... *My* intuition
doesn't seem to care that this doesn't resolve which sock in each pair
is which; the whole thing still 'feels like' a countable collection of
socks. As for (2), my intuition actually rebels a bit at infinite,
Dedekind-finite sets: the idea that for each n I can enumerate a
subset {x_0, x_1, ..., x_n} and yet can't enumerate an infinite subset
seems very counter-intuitive.

Brian M. Scott

[A Boucher]

In article <341a4218...@news.csuohio.edu> Brian M. Scott,

sc...@math.csuohio.edu writes:
>If I've understood you correctly, S = {s}, which is certainly a very
>special case. Your explanation of why you see no problem in picking
>an element from S doesn't obviously extend to finite sets of
>cardinality greater than 1.

Sure it does. Suppose S has two elements. Then we can "pick out" the
two elements by writing
"AxAy(x is in S and y is in S and x not equal to y ==> Bx and By)".

>I don't understand what makes it different. The choices are
>independent, so you may consider making them in parallel (i.e.,
>simultaneously). Why is making infinitely many choices more dubious
>than making A(N, N) choices, where A is the Ackerman function and N is
>the number of elementary particles in the known universe?

For me there is a big difference between N and infinity. For one thing,
one can represent the former with a first-order formula.

>>I would justify the axiom "0 is a natural number"
>>by saying it is intuitively obvious.

>It seems to me to be purely a matter of definition.

Okay, fine, but you agree then AC is not "a matter of definition".

>>I would justify other axioms (e.g.
>>whichever one we use to assume the existence of irrational numbers) by
>>saying they are useful.

>But you don't accept a justification of AC on the grounds that it's
>useful?

Clearly AC is useful to prove all the fun theorems that mathematicians
have proven these past hundred or so years. But are these theorems
useful, in the sense that they are vital? (See previous post.) I would
indeed accept AC (with some remorse!) if someone could explain to me it
is vital. Otherwise I feel entitled to use Occam's Razor...

[A Boucher]

In article <5vdd3s$j...@gap.cco.caltech.edu> Ilias Kastanas,
ika...@alumni.caltech.edu writes:

>In article <5vccb8$gkh$1...@belzebul.imaginet.fr>,
>A Boucher <abou...@imaginet.fr> wrote:
>>In article <5vbvej$f...@gap.cco.caltech.edu> Ilias Kastanas,
>>ika...@alumni.caltech.edu writes:
>>

>
> I understand; I'm trying to find out how this comes to terms with
>the fact that at each A, there is an x. That's infinitely many x, yes; why
>is that an obstacle to putting them in a set C? E.g. we say "let D be
>an infinite subset of [0,1]..." and we go ahead to prove it has a limit
>point. Well then, we just put infinitely many points into D. Why isn't
>_that_ objectionable?
>

In fact I do object to the proof that it has a limit point.

>
> I'm looking for some coherence. It is acceptable that an element
>can be picked out of each A; why would "the set of them" be problematic?

Because you must pick out an infinite number!

>
>

>>>In article <5v9prr$r09$1...@belzebul.imaginet.fr>,
>>>A Boucher <abou...@imaginet.fr> wrote:
>>>>As to the "so what", I'm asking what justifies AC. You answer lots of
>>>>mathematicians use it. Well, I'm not an expert in the history of
>>>>mathematics, but if Euler made a mistake, probably lots of mathematicians
>>>>at the same time made the same mistake. So lots of mathematicians can be
>>>>wrong about something. If about infinite series, why not about AC?
>>>
>>>
>>> Eh, this is not an argument against AC, it's against everything
>>>in mathematics! "Everybody could be wrong" about anything in particular...
>>
>>It's not an argument against everything in mathematics, because I would
>>never justify *any* axiom by appealing to the fact that most

>>mathematicians use it. I would justify the axiom "0 is a natural number"
>>by saying it is intuitively obvious. I would justify other axioms (e.g.

>>whichever one we use to assume the existence of irrational numbers) by
>>saying they are useful.
>
>
>

> The point is, mathematicians don't just use it, they also find it
>intuitively obvious. E.g. the four cases below.
>

> Usefulness. Would you accept something not intuitively obvious
>because it's useful? (And in what sense?)
>

I gave an example: the assumption of the existence of the irrational
numbers. Okay, it's not the level of sets, but consider axioms for real
analysis on the level of Apostol (p. 2-9). At some point you have to
make an assumption which will imply that irrational numbers exist. I
accept this, even though I do not think it is "intuitively obvious",
because the existence of irrationals seems to me a "vital" assumption, in
that it permeates the physicist's model of reality. AC does not cut it
for me in the same way, although I am open to being corrected. That is,
if mathematicians stopped using AC and had to add a few more assumptions
in their propositions (assumptions obeyed in any case by functions used
by physics, science, and everyday life), I think no one would notice,
except the mathematician.

By the way (back to a previous point), according to Morris Kline (p.
161, The Loss of Certainty), Ampere proved in 1806 that a function has a
derivative łat all points where it is continuous˛. łOther or similar
Śproofsą were given by Lacroix... and by almost all the leading texts of
the 19th century.˛

You just don't want to give me the credit of consistency. I will not use
a theorem which uses AC_w, unless I can prove it otherwise. So I used
choice when I learned calculus, so what? Russell in My Philosophical
Development: "Those who taught me the infinitesimal calculus did not
know the valid proofs of its fundamental therems and tried to persuade me
to accept the official sophistries as an act of faith."

>
> Anyway, since you seem to reject that (1) the union of countably many
>two-element sets is countable, or (2) any infinite set has a countable subset
>x_0, x_1, ... , I asked you to explain, if possible, what makes you see
>them as non-obvious or doubtful.
>

I'm really not sure how to reply. Sure, you can whack me over the head
if I were to claim, "Red is not a color" isn't obvious. But surely I'm
allowed to claim (1) and (2) are not obvious. Because really and truly,
they are not obvious to me. And really and truly, I would want a proof
of these propositions before accepting them. Do you find them so obvious
that you just accept them without proof!?

>
> Ilias
>

[Bill Taylor]

Fred Galvin <gal...@math.ukans.edu> writes:

|> Don't know if he does, but I do: "entities should not be multiplied
|> unnecessarily." And, since I accept Occam's Razor, I have to accept the
|> axiom of choice, because the axiom of choice (in its strongest form, a
|> definable well-ordering of the universe) follows from Goedel's axiom of
|> constructibility, and the axiom of constructibility is required by Occam's
|> Razor. If nonconstructible sets are not "unnecessary entities", please
|> tell me what they are needed for. Real math and science uses, not fancy
|> highbrow set-theory-for-set-theory's-sake.

Forgive me for this observation, Fred, but I think this is a bit ingenuous,
tempting thought it is as an argument.

Saying "every set is constructible" is NOT required in order to "ensure"
there are no unconstructible sets! Just leave it that you know there aren't
any, and forget about making it an axiom.

Because, the trouble with this axiom, (V=L), is that it *forces* a constructive
naming on to *every* set, whether it wants one or not! And some sets might
not; in fact do not. You are (in a concealed way) just as guilty of Occam's
heresy as using AC:- you are saying, "given any set, there is a specific way
of inserting it into a level of the L hierarchy", and this may be too
much to assume. IS too much too assume, in fact, because (as you have
almost noticed), V=L is just a form of choice - it merely says "every set
can be put into a well-ordered heierarchy"; NOT because it naturally goes
there, but BY FIAT.

This is clearly begging the whole question of AC.

MHO.

-------------------------------------------------------------------------------
Bill Taylor W.Ta...@math.canterbury.ac.nz
-------------------------------------------------------------------------------

Is aleph_69 a cardinal sin?
-------------------------------------------------------------------------------

[Brian M. Scott]

On 15 Sep 1997 19:20:32 GMT, A Boucher <abou...@imaginet.fr> wrote:

>In article <341a4218...@news.csuohio.edu> Brian M. Scott,
>sc...@math.csuohio.edu writes:
>>If I've understood you correctly, S = {s}, which is certainly a very
>>special case. Your explanation of why you see no problem in picking
>>an element from S doesn't obviously extend to finite sets of
>>cardinality greater than 1.

>Sure it does. Suppose S has two elements. Then we can "pick out" the
>two elements by writing
> "AxAy(x is in S and y is in S and x not equal to y ==> Bx and By)".

This allows you to assert something about the members of S, but I
don't see how it helps you to pick a member of S. I could almost
understand it if you objected to making even a single choice from a
set containing more than one element; what I don't understand is why
the number of arbitrary choices (finite vs. infinite) makes a
difference.

>>I don't understand what makes it different. The choices are
>>independent, so you may consider making them in parallel (i.e.,
>>simultaneously). Why is making infinitely many choices more dubious
>>than making A(N, N) choices, where A is the Ackerman function and N is
>>the number of elementary particles in the known universe?

>For me there is a big difference between N and infinity. For one thing,
>one can represent the former with a first-order formula.

I agree that there's a big difference between them. It just seemed to
me that in the everyday world the difference wasn't really very
important, unlike the difference between, say, 3 and A(N, N).

>>>I would justify the axiom "0 is a natural number"
>>>by saying it is intuitively obvious.

>>It seems to me to be purely a matter of definition.

>Okay, fine, but you agree then AC is not "a matter of definition".

I'm not sure that we're talking about the same thing. When you write
down a version of the Peano axioms that includes '0 is a natural
number', that axiom is part of a definition of 'natural number' and
'0'. You make the definition because it accords with your view of
these concepts. But given any reasonable definition of the natural
numbers, you could define a constant '0' with the usual meaning.

If I write down the axioms and axiom schemata of ZF or ZFC, I am
similarly defining a universe of sets. If I omit AC from my
description, it, unlike 0 in the previous example, is not implicitly
present. In that sense one is a matter of definition, and the other
isn't. One seems to me to be a statement about the meanings of
symbols; the other, a statement about the structure of the universe of
sets. The first needs no justification and therefore is not really
analogous to the second.

Let's go back to what I think you were trying to get at. Would you
justify the union axiom, Ax Ay Ez Au [u \in z <--> u \in x v u \in y],
by saying that it's intuitively obvious?

>>>I would justify other axioms (e.g.
>>>whichever one we use to assume the existence of irrational numbers) by
>>>saying they are useful.

>>But you don't accept a justification of AC on the grounds that it's
>>useful?

>Clearly AC is useful to prove all the fun theorems that mathematicians
>have proven these past hundred or so years. But are these theorems
>useful, in the sense that they are vital? (See previous post.)

I confess that I still don't understand why vitality (in your sense)
should have anything to do with truth.

>I would
>indeed accept AC (with some remorse!) if someone could explain to me it
>is vital. Otherwise I feel entitled to use Occam's Razor...

Well, here I will have to agree with Fred Galvin: Occam's Razor leads
to the conclusion that V = L, which implies AC. (I've never been
entirely comfortable with this shave, but that has nothing to do with
AC: I just don't trust the mathematical universe to be quite *that*
tame.)

Brian M. Scott

[Bill Taylor]

ika...@alumni.caltech.edu (Ilias Kastanas) writes:

|> The [parallel] Postulate? For the Euclidean plane, yes; for the hyperbolic,

|> no... The point is, it's not the same thing. There are many kinds of
|> geometries, or measures, or modules; but the operation of forming sets
|> appears to have something fundamental about it.

OUCH!

It's not like you to fall into such a historico-philosophical error, Ilias,
or even the appearance of one!

"The operation of forming sets appears to have something fundamental about it."

That's the sort of thing they used to say about commutativity of addition,
about the rationality of all magnitudes, about the fact that the numbers
stopped at 10. And, of course, about the fact of being able to draw a
unique parallel line through any point...

Hmmmmmmm...

-------------------------------------------------------------------------------
Bill Taylor W.Ta...@math.canterbury.ac.nz
-------------------------------------------------------------------------------

The Eternal Union of Soviet Republics lasted just
seven times longer than the Thousand Year Reich !
-------------------------------------------------------------------------------

[Bill Taylor]

wee...@sagi.wistar.upenn.edu (Matthew P Wiener) writes:

|> On this note, one should recall the original Banach-Tarski paradox paper
|> used AC for _two_ results: the infamous 3D paradox, of course, but also
|> 2D is _not_ similarly paradoxical.
|>
|> It was almost 20 years before a ZF proof of their 2D nonparadox result
|> was found.

Matts, can you please give us a brief outline of these two prooofs?

That is, the AC and the non-AC proofs of 2D-nonparadox.

Thanks IA.

-------------------------------------------------------------------------------
Bill Taylor W.Ta...@math.canterbury.ac.nz
-------------------------------------------------------------------------------

You are what you remember.
-------------------------------------------------------------------------------

[Bill Taylor]

ika...@alumni.caltech.edu (Ilias Kastanas) writes:

|> >Give that man a cigar! Mind you, its possible implausibility in the
|> >clumsy form Euclid felt obliged (by finitary considerations) to put it in,
|> >contributed greatly to a major revolution in math.

|> Eh... okay, so what is it, plausible or implausible?!

Sorry I didn't make it clear:- in Euclid's form it didn't seem so
unimpeachibly plausible, (hence the course of math history); but in
Playfair's fair play form it seems quite unimpeachibly and totally natural,
right and proper in every respect. Still seems that way to me... ;-)

|> One can see "finitary" leanings in "there is no greatest prime",

Yes, but elsewhere too. If you look at them closely, you'll find most of
his geometrical doings involve not just FINITE but also COMPACT figures.
All his lines are finite, and include their endpoints, etc etc.

The finiteness is clear from his postulate - it is possible to *extend*
any line in either direction. Clearly they are always finite.

|>(And it has been noticed that Euclid never used AC...)

After the trouble that Hippasus had with irrationals, he didn't dare!!

|> Even on a first reading, Euclid comes through clearly: one line
|> is parallel, and the others aren't.

IMHO this is a misinterpretation. He said there *is* a parallel. But not one,
indeed, by the above, it can always be extended. Euclid was unfortunately
*very* hazy in his attitude to whether a line (segnment) that included
another one was "the same one" or different. This haziness comes out most
strongly in his approach to SAS triangles:- "picking up" one line and
plonking it down over another. Very mucky!

And IMHO rather represhensible. He did geometry the way he did, to *avoid*
carving-up-and-replacement proofs, even when much simpler (e.g. Pythagoras,
whose simpler proof he a.s. knew). Then to turn around and use similar ideas
elsewhere, was very naughty. *AND* he knew it was, because he wemt out of his
way to avoid doing it when not necessary - he reduces other things to the SAS
case, even when another pick-up-and-plongk-down proof would be simpler.

Guilty conscience!

Bad karma for Euclid. (Why do you think I came back as a mere instructor? ;-)

|> manuscripts in a WWII soldier's backpack...

Eh? What is this story? Do tell!

|> I doubt Saccheri would do crosswords.

'Course not! He would be investigating the relationships between various
weak choice assumptions, like he shoulda done...

-------------------------------------------------------------------------------
Bill Taylor W.Ta...@math.canterbury.ac.nz
-------------------------------------------------------------------------------

The beauty of the Pythagorean theorem is hopelessly tainted by its
relevance to the approximately Euclidean space we actually inhabit.
--- John Baez (misquoted)
-------------------------------------------------------------------------------

[Fred Galvin]

On 15 Sep 1997, Bill Taylor wrote:

> That's about it. It could be said to be a case of ontological minimality,
> to put a pompous label on it, or Occam's razor, to give it a patina of
> scholastic respectability. But it just amounts to noticing that non-definable
> things "just don't exist", however consistent they may be.

You seem to be pretty consistent, Bill, but are you definable?

> On this general topic, there are some remarkably tricky things proved with AC
> in ordinary R^3. Things like, "it is possible to partition 3-D space into
> lines all with different slopes"; or "circles all with nonparallel planes";
> or other things like so. Alas, my colleague who told me about such matters
> many years ago can't recall them now!
>
> Can anyone else help?

I sure hope so, because the best I can do is mention a vaguely recollected
paper by Conway and Croft about geometric partitions which might or might
be related somehow to the things you're asking about. The paper I can't
remember much about might or might not be "Covering a sphere with
congruent great-circle arcs" in Proc. Cambridge Philos. Soc. 60 (1964),
787-800, which reference I found jotted down on a file card, but maybe
it's some other paper by the same authors, or was it two other guys.

[Ilias Kastanas]

In article <5vigga$il7$1...@cantuc.canterbury.ac.nz>,

Bill Taylor <mat...@math.canterbury.ac.nz> wrote:
>ika...@alumni.caltech.edu (Ilias Kastanas) writes:
>
>|> >and be prepared to look at distinctions in its absence.
>
>|> Well, Bill, you do know I don't hold back from such
>
>Indeed I do! You're a good guy though, Ilias, pity there aren't more of you.

Oh boy, imagine a few more around...

>|> You and others have such against AC, and I would like to
>|> hear what they are. There have been discussions in the past; what is it
>|> like in the present?

>Well, my motivation is simple enough. It is merely that the only sets

>which "really exist" are definable ones; (and quite possibly not even all
>of those! But that's a whole nother thread.)
>

>It seems this was Lebesgue's (& Borel's) idea also; but they lived back before
>such concepts could be neatly codified, (as e.g. in my w_1^CK post, though
>others have done better).

Well, the issue here is that (as Fred Galvin has also mentioned)
if we only "accept" what is definable, then Wellordering holds -- in fact
there is a global wellordering of the universe. This is certainly so if
"definable" means first-order (or anything less); we end up with the con-
structible sets (L).

Now L is hardly abstruse or unreal. In fact, one only need
look at it via Goedel's older approach: all we are doing is forming sets
like {x,y} or x-y or dom(x) or permuting the triples in x. Even if one
considers that proceeding "throughout the ordinals" is unclear, the fact is
the wellordering is built along the way; we have it, "wherever" we wish to
stop!

One might relax "definable" into including ordinal parameters (the
ordinal-definable sets OD, and HOD); but that global ID-issuer is still there.

By the way, remember the "shotgun marriage" of Wellordering (qua ORD,
and transfinite induction over it) and Powerset? It's amusing to note that
Powerset implies arbitrarily large cardinals, and hence ORD and class-length
t.i.'s; conversely, t.i. over ORD builds L via "definable" Powerset... and
ends up implying Powerset, since the latter holds formally in L!

>So all those lovely choice-necessitous things - Hamel bases, nonmeasurable
>sets, Banach-T partitions, unprincipled infiltrators, etc etc, which ARE
>lovely things, no question, simply "aren't there" (in ontological space!)

>So results about them, and many results using them, are more like "what it
>would be like if these (impossible) things *did* exist". A kind of fantasy
>math. What was Weemba's term for it...? Oh yes, "SURREAL math". Nice.

But we have already indulged in the "surreal" in accepting the set
of all reals; surely only some of them can be definable. And every w-sequence
of reals is coded by a real, so we have those too. Isn't it "too late" to
insist that when it comes to sets of reals, or functions to/from R, P(R), ...
only "definable" ones can be there? After having admitted R, P(R),... lock,
stock and barrel? Note also that there can be sets of reals that are
definable without even a single real in them being definable.

>Not that I've anything against it at all. It's all legitimate math, math
>doesn't *have* to be fully "real", as we know. I'm all for Hilbert and
>the pure math toast! But still, it's irritating to see surreal math
>passed off as real math, as is automatically assumed in today's texts. Well,
>perhaps not quite automatically... was it Kreisel who said, "AC is unique
>in its ability to trouble the conscience of the ordinary mathematician!"

It is informative to be able to produce a choice function explicitly
rather than by appeal to AC; what I'm saying is that once we accept infinite
sets the way we have, AC follows from the same intuition, not an extension.
We collect "all subsets of w" to form R, without giving an effective
procedure that will produce them, one after another; we take any one that is
there. So if some real is there in each A of a family F, it seems we ought
to be able to do the same, without a specifying procedure. It's not formally
provable, of course, but it is based on the same idea.

One can work in ZF, or in various ZF + X - Y; what I'm saying is
that ZF + AC has the same "surreality" as ZF, from an intuitive viewpoint.

>AC_w being used in 19th-century style math is not nearly so heinous; most
>such cases (of existence proofs), can easily be made constructive for any
>particular example. It's very hard to drum up *really* nonconstructive
>things on the real line; the early intuitionists notwithstanding. But no

Hmm... like, every continuous positive function on [0,1] has a
positive l.u.b., but we may not assert it occurs at a definite x...

>doubt it could be done, with extraordinary effort; it just won't happen in
>bridge-building math. And anyway, there's always the conservative extension
>meta-theorem to fall back on!

I'm not clear as to which metatheorem you mean.

One can eliminate AC_w (and AC too) in a number of cases... You
don't need them, e.g., to prove Heine-Borel for [0,1], or that R is not
meager. But getting by without having that R is not a countable union of
meager sets (or just countable ones), without countable additivity of Le-
besgue measure,... , proving them ad hoc in cases you can... ouch!

Another thing. Sure, AC_w is weaker than AC; it's the rule,
after all, in the whole realm of weak choice, or medium or medium-rare...
But the view that AC_w is intuitively "more" justifiable, maybe to the
point of granting AC_w while denying AC, is something I don't get. (I
don't know whether you hold it; some people do). What basic intuition or
principle entails that an infinite F should have a choice function if it is
countable but not otherwise? It sounds like the outcome of just being more
familiar and comfortable with countable sets than other infinite ones.

>So.

>That's about it. It could be said to be a case of ontological minimality,
>to put a pompous label on it, or Occam's razor, to give it a patina of
>scholastic respectability. But it just amounts to noticing that non-definable
>things "just don't exist", however consistent they may be.

Okay, Occam's razor cuts both ways... minimality can be taken all
the way to L_beta_0, and beta_0 is countable, though much larger than w_1-CK.
Again, with only definable things we end up with a global wellordering (de-
finable! -- Sigma-1-in ZF), and hence with a definable wellordering of R
(Sigma-1-2); Hamel bases, nonmeasurable sets and other thingies are all
there, card-carrying definable citizens! Is there some tinkering with
definability that alters this pattern?

A stray note: under AC every set in a model has a bijection with
an ordinal, of course; it follows that every set can be encoded into a set
of ordinals. Thus any model is _determined_ by its sets of ordinals; so
in a sense AC says that "arbitrary" sets do not exist... all there is is
subsets of ORD! How's that for a reversal? AC actually makes all sets be
of a rather concrete kind!

>On this general topic, there are some remarkably tricky things proved with AC
>in ordinary R^3. Things like, "it is possible to partition 3-D space into
>lines all with different slopes"; or "circles all with nonparallel planes";
>or other things like so. Alas, my colleague who told me about such matters
>many years ago can't recall them now!
>
>Can anyone else help?

Aha... We can have a quadric that consists of straight lines all
with different slopes; AC just lifts this to R^3. How natural!

>-------------------------------------------------------------------------------
> Bill Taylor W.Ta...@math.canterbury.ac.nz
>-------------------------------------------------------------------------------

> There are no objects such that there are no such objects.
>-------------------------------------------------------------------------------

Eh... such a statement is well-liked by people who would like
this sort of statement.

And offhand, partition-into-lines looks like a transfinite induction
similar to, uh, those used for this kind of thing. Suppose lines L_0, ...
L_a are pairwise non-intersecting (at infinity(!) or otherwise), for an
ordinal a < c = |R|. Let Q be the least (in the wellordering of R^3) point
not on any L_i. There is a plane P through Q distinct from the plane through
Q and L_i, for all i <= a (the least in the wellordering of planes, natch).
Each L_i intersects P at one point F_i (at infinity or not); let L be a line
on P through Q, distinct from the a lines QF_i. Define L_a+1 to be L.
(Similarly if the original lines are L_i, i < a, a limit; L becomes L_a).

For less than c -many L_i, there will always be a Q (easy... though
"m(L_i) = 0, hence m(union) = 0" is not necessarily correct). But we need
to be sure L_i, i < c does exhaust R^3 ! That's easy, too; as long as
R^3 is wellordered in the minimal type c... since the first j points are in
the union of the first j L_i 's.

Yes, yes, a good illustration of the evil nature of AC, right?!

Ilias

[Ilias Kastanas]

In article <5vk43o$mt9$1...@belzebul.imaginet.fr>,

A Boucher <abou...@imaginet.fr> wrote:
>In article <5vdd3s$j...@gap.cco.caltech.edu> Ilias Kastanas,
>ika...@alumni.caltech.edu writes:
>>
>> I understand; I'm trying to find out how this comes to terms with
>>the fact that at each A, there is an x. That's infinitely many x, yes; why
>>is that an obstacle to putting them in a set C? E.g. we say "let D be
>>an infinite subset of [0,1]..." and we go ahead to prove it has a limit
>>point. Well then, we just put infinitely many points into D. Why isn't
>>_that_ objectionable?
>>
>
>In fact I do object to the proof that it has a limit point.
>
>>
>> I'm looking for some coherence. It is acceptable that an element
>>can be picked out of each A; why would "the set of them" be problematic?
>
>Because you must pick out an infinite number!

The limit point proof does not involve choice; you seem to be
objecting to just forming D... or maybe infinite sets in general.

...

>> The point is, mathematicians don't just use it, they also find it
>>intuitively obvious. E.g. the four cases below.
>>
>> Usefulness. Would you accept something not intuitively obvious
>>because it's useful? (And in what sense?)
>>
>
>I gave an example: the assumption of the existence of the irrational
>numbers. Okay, it's not the level of sets, but consider axioms for real
>analysis on the level of Apostol (p. 2-9). At some point you have to
>make an assumption which will imply that irrational numbers exist. I
>accept this, even though I do not think it is "intuitively obvious",
>because the existence of irrationals seems to me a "vital" assumption, in
>that it permeates the physicist's model of reality. AC does not cut it
>for me in the same way, although I am open to being corrected. That is,
>if mathematicians stopped using AC and had to add a few more assumptions
>in their propositions (assumptions obeyed in any case by functions used
>by physics, science, and everyday life), I think no one would notice,
>except the mathematician.

Again, there is no AC in forming R (basically, it is "the sets of
integers", P(N))... but we do put infinitely many things into R, and you
don't think that's obvious either. Maybe N, too... collecting infinitely
many integers?!

Anyway, I was asking about fundamental principles; "I accept what
is vital for a physicist's model" isn't one... and it's hardly relevant or
interesting. No, I don't disparage physics; I like it, I've done some.
But this is putting the cart before the horse. One needs the properties
of a math object before deciding wheher it's appropriate for some use.
By the way, the Lebesgue integral and L^2 as Hilbert space are important
in physics; measure theory uses AC, and so does the Hahn-Banach theorem,
uniform boundedness, etc. But it doesn't matter; physics is not the
criterion.

>By the way (back to a previous point), according to Morris Kline (p.
>161, The Loss of Certainty), Ampere proved in 1806 that a function has a
derivative łat all points where it is continuous˛. łOther or similar
>Śproofsą were given by Lacroix... and by almost all the leading texts of
>the 19th century.˛
>

Ampere was not exactly a mathematician... Did Jordan's "Cours
d' analyse" give such a proof?

No offense meant; I'm just pointing out consistency in this matter
is no simple thing, i.e. sifting through all theorems and making sure not
to use those that need some form of AC,

>a theorem which uses AC_w, unless I can prove it otherwise. So I used
>choice when I learned calculus, so what? Russell in My Philosophical
>Development: "Those who taught me the infinitesimal calculus did not
>know the valid proofs of its fundamental therems and tried to persuade me
>to accept the official sophistries as an act of faith."

That's different. He was given wrong proofs, while correct ones
were possible. Here we have proofs that are correct, granting AC_w or AC,
and in most cases it is known there is no proof without AC(w)... provably so!

So are you actually _not_ using calculus results like above, or
any consequence of theirs?

>> Anyway, since you seem to reject that (1) the union of countably many
>>two-element sets is countable, or (2) any infinite set has a countable subset
>>x_0, x_1, ... , I asked you to explain, if possible, what makes you see
>>them as non-obvious or doubtful.
>>
>
>I'm really not sure how to reply. Sure, you can whack me over the head
>if I were to claim, "Red is not a color" isn't obvious. But surely I'm
>allowed to claim (1) and (2) are not obvious. Because really and truly,
>they are not obvious to me. And really and truly, I would want a proof
>of these propositions before accepting them. Do you find them so obvious
>that you just accept them without proof!?

To prove (1), let f_n enumerate the n-th set, i.e. if A_n = {a, b}
then f_n(0) = a, f_n(1) = b. You define g by g(2n) = f_n(0), g(2n +1) =
= f_n(1), and g then is an enumeration of the union of the A_n.

This does seem simple and straightforward, right?

Well, it uses AC_w. The assumption was that each A is a pair, i.e.
for each A there is an enumeration f of domain {0,1}... in fact, exactly
two such enumerations. But to put together f_n we need to pick one such
enumeration f for each n... so we invoke AC_w, and then set f_n(x) = f(x).

OK, maybe we can find a way to do it without using AC_w...?! No.
There is a model of mathematics, satisfying all other axioms, where there
are A_n's, each having its two f's, but there is no f_n and no g.

Now if you are saying we cannot form f_n, and in fact we cannot
form any infinite set other than explicit ones like { 3n^2 +2 : n in N},
-- unless if we encounter something in physics that makes us like the
idea of having an f_n, in which case it does exist... as long as we don't
change our mind... well, it's a view, but it has nothing to do with
mathematics.

Ilias

Ilias
>>

[Ilias Kastanas]

In article <5vl2k1$khq$3...@cantuc.canterbury.ac.nz>,

Bill Taylor <mat...@math.canterbury.ac.nz> wrote:
>
>ika...@alumni.caltech.edu (Ilias Kastanas) writes:
>

>|> The [parallel] Postulate? For the Euclidean plane, yes; for the hyperbolic,

>|> no... The point is, it's not the same thing. There are many kinds of
>|> geometries, or measures, or modules; but the operation of forming sets
>|> appears to have something fundamental about it.
>
>

>OUCH!
>
>It's not like you to fall into such a historico-philosophical error, Ilias,
>or even the appearance of one!
>
>"The operation of forming sets appears to have something fundamental about it."
>
>That's the sort of thing they used to say about commutativity of addition,
>about the rationality of all magnitudes, about the fact that the numbers
>stopped at 10. And, of course, about the fact of being able to draw a
>unique parallel line through any point...
>
>Hmmmmmmm...

It's nice of you, Bill, to insert a compliment even in criticism!
Now, am I deserving ... of either?!

I don't know who may have said commutativity of + is fundamental...
Anyway, at a time when + referred to N, or Q, that is correct; it is a
fundamental property. When people envisioned displacements or rotations
and used '+' on them, it turned out these new +'s sometimes were commutative
and sometimes not. Well, so what? They happened to use the same symbol;
that's in no way a downfall of some prior certainty! I fail to see anything,
eh, fundamental here... In the context I was writing above I would just
say, there are many 'additions'...

Likewise, it is and has been well known there are number systems
other than base 10. And for the parallel, the usual notion of distance
makes it Euclidean. People may have missed that the hyperbolic plane, and
distance, are possible. Call it a mistake; it is just about the geometry
of one thing, the plane. It's pretty clear a different geometry is going
on on a sphere.

And really, that others may have said something similar to what I
said, in this or that setting, leaves me cold.

Some mathematical objects, rings or topologies or measures, are
multifaced -- intentionally, too. Others have a good claim of uniqueness,
like R... or an overwhelming one, like w. I'm saying sethood has such a
claim (and w and R seem partial cases of it). Wellfounded sets based
on 0 (= {}) come in orderly stages. When forming P(w), maybe we cannot
describe in detail each and every member... maybe you or somebody wants to
suggest a limitation to definable ones... But it is clear what the general
notion "subset of w" is, and if we use it there is a strong suggestion we
are putting into P(w) "all there is", that there is a maximal collection.
We are exhausting all possibilities -- and likewise for every object and
every level.

So I'm claiming there is a strong suggestion of "something funda-
mental about it". It is not certain, no. Since Goedel voiced his belief
set theory has developed a lot, and results seem mixed in providing support
or refutation. But it seems in no way less strong today than back then.

Ilias

[Ilias Kastanas]

In article <5vl266$khq$1...@cantuc.canterbury.ac.nz>,

Bill Taylor <mat...@math.canterbury.ac.nz> wrote:
>Fred Galvin <gal...@math.ukans.edu> writes:
>
>|> Don't know if he does, but I do: "entities should not be multiplied
>|> unnecessarily." And, since I accept Occam's Razor, I have to accept the
>|> axiom of choice, because the axiom of choice (in its strongest form, a
>|> definable well-ordering of the universe) follows from Goedel's axiom of
>|> constructibility, and the axiom of constructibility is required by Occam's
>|> Razor. If nonconstructible sets are not "unnecessary entities", please
>|> tell me what they are needed for. Real math and science uses, not fancy
>|> highbrow set-theory-for-set-theory's-sake.
>
>Forgive me for this observation, Fred, but I think this is a bit ingenuous,
>tempting thought it is as an argument.
>
>Saying "every set is constructible" is NOT required in order to "ensure"
>there are no unconstructible sets! Just leave it that you know there aren't
>any, and forget about making it an axiom.

But Bill, if we decide there are no undefinable or non-constructible
sets, that does mean every set is constructible... how can we forget about
it? Sets are a certain way... but we don't look at what that means?!

>Because, the trouble with this axiom, (V=L), is that it *forces* a constructive
>naming on to *every* set, whether it wants one or not! And some sets might
>not; in fact do not. You are (in a concealed way) just as guilty of Occam's
>heresy as using AC:- you are saying, "given any set, there is a specific way
>of inserting it into a level of the L hierarchy", and this may be too
>much to assume. IS too much too assume, in fact, because (as you have
>almost noticed), V=L is just a form of choice - it merely says "every set
>can be put into a well-ordered heierarchy"; NOT because it naturally goes
>there, but BY FIAT.
>
>This is clearly begging the whole question of AC.
>
>MHO.
>

>-------------------------------------------------------------------------------
> Bill Taylor W.Ta...@math.canterbury.ac.nz
>-------------------------------------------------------------------------------

> Is aleph_69 a cardinal sin?
>-------------------------------------------------------------------------------

Is it really begging the question?

L is not beholden to V = L; it is given by a class definition in
ZF, and can be carried out in any model of it U to yield L^U. One point
is absoluteness: we get the same things at each step, whatever the U. Those
are sets that have to be there!

So we are not bringing in an evil, wellordered hierarchy and imposing
it, from outside. It is just that L is naturally wellordered... as "definable"
things are wont to be. Every set is definable; so it has definable elements
too; so do they; etc When we carry this out we get L, with a definable
(class) wellordering.

Since L turns out to be a model of ZF, it becomes _possible_ to
say V = L; (it could have been that definable sets are not sufficient to
produce a model). Moreover, V = L turns out to hold in L, too. If we
construct L^L, we don't get something less; we get all of L.

So if we accept every set is constructible, all this is inevitable.
We have L! We are not arbitrarily assigning 'names' either; a set's name
is its defining formula... and we have assumed such exists! It is not by
fiat that a poor arbitrary set is dragged willy-nilly and relegated to L;
it is that our assumption implies the only sets that exist _are_ those in
L. Once we make the assumption, we have no further choice... and we end
up with Choice! (Not to mention GCH... and Diamond... and other wonders!)

Ilias

[Fred Galvin]

On 16 Sep 1997, Bill Taylor wrote:

> Fred Galvin <gal...@math.ukans.edu> writes:
>
> |> Don't know if he does, but I do: "entities should not be multiplied
> |> unnecessarily." And, since I accept Occam's Razor, I have to accept the
> |> axiom of choice, because the axiom of choice (in its strongest form, a
> |> definable well-ordering of the universe) follows from Goedel's axiom of
> |> constructibility, and the axiom of constructibility is required by Occam's
> |> Razor. If nonconstructible sets are not "unnecessary entities", please
> |> tell me what they are needed for. Real math and science uses, not fancy
> |> highbrow set-theory-for-set-theory's-sake.
>
> Forgive me for this observation, Fred, but I think this is a bit ingenuous,

Not sure what you mean by "ingenuous". My dictionary says: ingenious;
noble, honorable; showing innocent or childlike simplicity and candidness;
lacking craft or subtlety; syn see natural. Hmm, maybe I should crosspost
this to alt.usage.english.

> tempting thought it is as an argument.
>
> Saying "every set is constructible" is NOT required in order to "ensure"
> there are no unconstructible sets! Just leave it that you know there aren't
> any, and forget about making it an axiom.

Say what? Would you mind going over that one more time? If we *know*
something about sets, why in hell shouldn't we make it an axiom? Isn't
that what axioms are for?

But, unlike you, I don't "know" that there are no nonconstructible sets.
Maybe they really do "exist" in some metaphysical sense, as ideas in the
mind of God, or some such thing. (The plausibility arguments in their
favor, at least the ones I've heard, do seem a lot more far-fetched and
less convincing than the ones for AC.) However, we Occamists regard
theoretical entities, such as sets, as a necessary evil, to be used as
sparingly as possible. It would be desirable to dispense with sets
entirely, since they don't really exist in the physical world. If we can't
get along without sets, then let's have as few of them as we can and still
have a smooth-running theory. Thanks to Goedel, we know that the
constructible sets are enough for all practical purposes. Therefore, we
Occamists only use the constructible sets (and damned few of those, but
that's beside the point), and we see no need to discuss any other kind.
Since constructible sets are the only ones we talk about, we just call
them "sets" (so you will have to find some other name for those
nonconstructible fantasies), and yes, every set has a well-ordering.

Some anti-Occamists, on the other hand, are enamored of sets, can't have
enough of them. Some of them, I think, believe that Cantor's Absolute
really exists: that God created an immense universe of sets, and it is our
religious duty to try to learn as much about it as we can. The infidel
set-theorists just think set theory is fun, and they want to have lots of
sets to do surreal math with.

It seems to me that, if you're going to admit *some* nonconstructible
sets, you should let them *all*. To accept some nonconstructible sets and
then not admit their choice functions or their well-orderings seems kind
of arbitrary. If you want a choiceless universe, you've got to pick and
choose which nonconstructible sets you let in. There is no practical or
moral or philosophical reason to do anything like that. The only reason I
can see is if you want to have fun with choiceless set theory, e.g., the
Dedekind-finite cardinals as a nonstandard model of arithmetic.

> Because, the trouble with this axiom, (V=L), is that it *forces* a
> constructive naming on to *every* set, whether it wants one or not! And
> some sets might not; in fact do not. You are (in a concealed way) just
> as guilty of Occam's heresy as using AC:- you are saying, "given any
> set, there is a specific way of inserting it into a level of the L

Nope, no insertions. Either it's there already, or it stays out.

> hierarchy", and this may be too much to assume. IS too much too assume,
> in fact, because (as you have almost noticed), V=L is just a form of
> choice - it merely says "every set can be put into a well-ordered
> heierarchy"; NOT because it naturally goes there, but BY FIAT.

It says nothing of the kind. All it says is that nonconstructible sets are
relegated to oblivion. As it happens, the ones that remain are provably
well-ordered, but that was not the criterion, it just worked out that way.

[A Boucher]

In article <5vlqhm$d...@gap.cco.caltech.edu> Ilias Kastanas,
ika...@alumni.caltech.edu writes:

>Subject: Re: WEAK CHOICE - another query.
>From: Ilias Kastanas, ika...@alumni.caltech.edu
>Date: 16 Sep 1997 11:28:54 GMT

>>In article <5vk43o$mt9$1...@belzebul.imaginet.fr>,
>A Boucher <abou...@imaginet.fr> wrote:
>>In article <5vdd3s$j...@gap.cco.caltech.edu> Ilias Kastanas,
>>ika...@alumni.caltech.edu writes:
>>>
>>> I understand; I'm trying to find out how this comes to terms with
>>>the fact that at each A, there is an x. That's infinitely many x, yes; why
>>>is that an obstacle to putting them in a set C? E.g. we say "let D be
>>>an infinite subset of [0,1]..." and we go ahead to prove it has a limit
>>>point. Well then, we just put infinitely many points into D. Why isn't
>>>_that_ objectionable?
>>>
>>
>>In fact I do object to the proof that it has a limit point.
>>
>>>
>>> I'm looking for some coherence. It is acceptable that an element
>>>can be picked out of each A; why would "the set of them" be problematic?
>>
>>Because you must pick out an infinite number!
>
>
> The limit point proof does not involve choice; you seem to be
>objecting to just forming D... or maybe infinite sets in general.
>

No, it's a little more subtle than that. You do need a version of AC in
one of its common proofs: look at both half-intervals, pick an element
of one which is non-empty, look at its half-intervals, etc. Granted you
don't need AC if you use the lub principle.

>
>...
>>> The point is, mathematicians don't just use it, they also find it
>>>intuitively obvious. E.g. the four cases below.
>>>
>>> Usefulness. Would you accept something not intuitively obvious
>>>because it's useful? (And in what sense?)
>>>
>>
>>I gave an example: the assumption of the existence of the irrational
>>numbers. Okay, it's not the level of sets, but consider axioms for real
>>analysis on the level of Apostol (p. 2-9). At some point you have to
>>make an assumption which will imply that irrational numbers exist. I
>>accept this, even though I do not think it is "intuitively obvious",
>>because the existence of irrationals seems to me a "vital" assumption, in
>>that it permeates the physicist's model of reality. AC does not cut it
>>for me in the same way, although I am open to being corrected. That is,
>>if mathematicians stopped using AC and had to add a few more assumptions
>>in their propositions (assumptions obeyed in any case by functions used
>>by physics, science, and everyday life), I think no one would notice,
>>except the mathematician.
>
>
>
> Again, there is no AC in forming R (basically, it is "the sets of
>integers", P(N))... but we do put infinitely many things into R, and you
>don't think that's obvious either. Maybe N, too... collecting infinitely
>many integers?!
>

Did I say AC is used to form R? Where???? You asked me for an example
of an axiom which I did not consider intuitively obvious but which I
accepted because it was useful. I gave you an example. I'll give you
another: Euclid's Parallel Line Postulate. And I don't mean to say that
the Postulate has anything to do with AC!

> Anyway, I was asking about fundamental principles; "I accept what
>is vital for a physicist's model" isn't one... and it's hardly relevant or
>interesting.

(1) Excuse me, but you cannot claim to be one interested in fundamental
principles when you seriously cite as a justification for AC that most
current mathematicians use the thing. What is fundamental about that?
(2) Sure it's relevant or interesting. It has to do with why I think we
should do mathematics.

>No, I don't disparage physics; I like it, I've done some.
>But this is putting the cart before the horse. One needs the properties
>of a math object before deciding wheher it's appropriate for some use.

>By the way, the Lebesgue integral and L^2 as Hilbert space are important
>in physics; measure theory uses AC, and so does the Hahn-Banach theorem,
>uniform boundedness, etc.

Now you're responding in the way I expected. An improvement!

>But it doesn't matter; physics is not the
>criterion.
>

There goes the improvement!

A simple proof, okay. But does that make it obvious? Just to summarize
the ridiculousness of it all: You claim that (1) is obvious. I claim it
is not. You reply with a simple and straightfoward *proof*. There is a
difference between an obvious proposition and a simple proof. No? That
there are an infinite number of primes admits a simple proof, but the
proposition is not obvious.

>
> Now if you are saying we cannot form f_n, and in fact we cannot
>form any infinite set other than explicit ones like { 3n^2 +2 : n in N},
>-- unless if we encounter something in physics that makes us like the
>idea of having an f_n, in which case it does exist... as long as we don't
>change our mind...

>well, it's a view, but it has nothing to do with
>mathematics.
>

Well la-di-dah, the gods have spoken. Some other viewpoints: "The value
or significance of the results deduced from the axioms was judged, at
least until severty-five or so years ago, by what they affirmed about the
physical world." (Kline) "The only consistent point of view...is rather
the assumption that the source and last raison d'etre of the number
concept--not only the natural but also the real
numbers--lie in experience and practical applicability." (Mostowski)
"If one was willing to accept the sciences, one might as well accept the
classical system of mathematics." (von Neumann) "How does Peano arrive
at his own fundamental principles...since after all he cannot prove them
either? Evidently by analyzing the modes of inference that in the course
of history came to be recognized as valid and by pointing out that the
principles are intuitively evident and necessary for science." (Zermelo)

Let's give a summary...

You asked for grounds on which AC "might be questionable".

I reply you are the one who should provide the grounds. By Occam's
Razor, it is the person who is accepting, not the person who is agnostic,
who should supply the reasons. You make a joke about Occam's Razor. Ha
ha.

I present two possibilities for accepting an axiom: (1) it is
intuitively obvious; (2) it is useful in some way.

You ignore (2)--which, accept my advice, is clearly the way to go--and
instead you say that the AC is intuitively obvious. Well, not quite, AC
is "natural and plausible". Now most people wouldn't say that
intuitively obvious is the same as natural and plausible, but apparently
that's not a problem.

At the end of the day, you have provided two reasons to accept AC:
(1) It is natural and plausible;
(2) Most mathematicians accept it.

(2) is clearly (may I say obviously?) wrong because mathematicians have
been wrong before. You (or someone else, excuse me if my memory fails
me) think that doesn't matter, because mathematicians *cannot* be wrong
in this case, because AC is independant of the other axioms. But that
isn't the conclusion you need; the conclusion you need is that we should
accept AC. And so the argument can't be right, because there exists x
such that most mathematicians have accepted x in the past, yet we should
not accept x. (e.g. take x = all continuous functions are almost
everywhere differentiable). You can come back and say your argument is:
if most mathematicians accept x and x is independant of the other
axioms, then we should accept x. But that dosn't hold either; take x =
Euclid's Parallel Line Postulate, which you yourself admit we should not
accept if we want to do a certain kind of non-Euclidean geometry.

So that leaves (1). Truth to say, I do not find AC natural and
plausible, but you have set the bar so low, I will grant you the
possibility that *you* can find it natural and plausible. But that
wasn't the original point, which is why we might find AC questionable.
If I can in good faith find AC is not natural and plausible, and you have
no other justificiation than (1), then AC is questionable.

In any case, we're arriving full circle and not really advancing. I
suggest you make one more response (if you want), and then we drop it!

[Brian M. Scott]

On 15 Sep 1997 05:19:06 GMT, mat...@math.canterbury.ac.nz (Bill
Taylor) wrote:

>On this general topic, there are some remarkably tricky things proved with AC
>in ordinary R^3. Things like, "it is possible to partition 3-D space into
>lines all with different slopes"; or "circles all with nonparallel planes";
>or other things like so. Alas, my colleague who told me about such matters
>many years ago can't recall them now!
>
>Can anyone else help?

The first one's not too hard; I haven't thought about the second, but
I suspect that it's similar. Well-order R^3 in type 2^w. At stage k
you want to choose a line L(k) though x(k) that has a different slope
from and is disjoint from each L(m), m < k. (This is assuming that
x(k) isn't already in the union of the L(m) already chosen; in that
case there's nothing to do at this stage.)

Each L(m), m < k, determines and is contained in a plane through x(k).
Since there are fewer than 2^w lines L(m), there is a plane, P, such
that (1) P contains x(k), and (2) for each m < k, P is not parallel to
(and therefore in particular doesn't contain) L(m). (For instance,
let S be the unit sphere - not ball - about x(k). For each m < k let
A(m) be the 2-point intersection of S with the line through x(k)
parallel to L(m). Now pick any point y on S that's not in the union
of the A(m), and let P contain x(k) and y.) Each L(m) meets P in a
single point. There are fewer than 2^w such points, so there's a line
in P containing x(k) that misses all of the L(m), m < k; choose such a
line, and let it be L(k).

Brian M. Scott

[Ilias Kastanas]

In article <5vmtea$aai$1...@belzebul.imaginet.fr>,

A Boucher <abou...@imaginet.fr> wrote:
>In article <5vlqhm$d...@gap.cco.caltech.edu> Ilias Kastanas,
>ika...@alumni.caltech.edu writes:

...

>>>>is that an obstacle to putting them in a set C? E.g. we say "let D be
>>>>an infinite subset of [0,1]..." and we go ahead to prove it has a limit
>>>>point. Well then, we just put infinitely many points into D. Why isn't
>>>>_that_ objectionable?
>>>>
>>>
>>>In fact I do object to the proof that it has a limit point.
>>>
>>>>
>>>> I'm looking for some coherence. It is acceptable that an element
>>>>can be picked out of each A; why would "the set of them" be problematic?
>>>
>>>Because you must pick out an infinite number!
>>
>>
>> The limit point proof does not involve choice; you seem to be
>>objecting to just forming D... or maybe infinite sets in general.
>>

>No, it's a little more subtle than that. You do need a version of AC in
>one of its common proofs: look at both half-intervals, pick an element
>of one which is non-empty, look at its half-intervals, etc. Granted you
>don't need AC if you use the lub principle.

Not quite; D might have only finitely many points in a half-interval.

You don't want to pick elements of D; you pick half-intervals. Na-
mely, you pick the left one iff D has infinitely many points in it. No
AC neded for this. The intersection of this sequence of half-intervals is
a single point, again without AC; and that's the limit point.

Did I say you said so?

On the contrary -- I'm pointing out you are objecting to the forming
of infinite sets, even in cases AC is not involved. So it's not AC per se;
you are also rejecting parts of classical mathematics without AC.

>> Anyway, I was asking about fundamental principles; "I accept what
>>is vital for a physicist's model" isn't one... and it's hardly relevant or
>>interesting.
>
>(1) Excuse me, but you cannot claim to be one interested in fundamental
>principles when you seriously cite as a justification for AC that most
>current mathematicians use the thing. What is fundamental about that?

The fact that they use it _because_ they consider it a fundamental
principle.

>(2) Sure it's relevant or interesting. It has to do with why I think we
>should do mathematics.

That makes it interesting for you. Others may think otherwise.

>>No, I don't disparage physics; I like it, I've done some.
>>But this is putting the cart before the horse. One needs the properties
>>of a math object before deciding wheher it's appropriate for some use.
>
>>By the way, the Lebesgue integral and L^2 as Hilbert space are important
>>in physics; measure theory uses AC, and so does the Hahn-Banach theorem,
>>uniform boundedness, etc.
>
>Now you're responding in the way I expected. An improvement!
>
>>But it doesn't matter; physics is not the
>>criterion.
>
>There goes the improvement!

No "improvement"; I mentioned them since you wanted to know. It's
not an argument for or against AC.

Never mind the wordplay. I gave a simple _and_ obvious proof of
(1)... unlike the simple proof for the primes, which takes ingenuity to
find. And I pointed out, the proof does use AC_w, and most people do find
the whole thing obvious and convincing. You snipped that part, to try to
pretend the matter is ridiculous.

>> Now if you are saying we cannot form f_n, and in fact we cannot
>>form any infinite set other than explicit ones like { 3n^2 +2 : n in N},
>>-- unless if we encounter something in physics that makes us like the
>>idea of having an f_n, in which case it does exist... as long as we don't
>>change our mind...
>
>>well, it's a view, but it has nothing to do with
>>mathematics.

>Well la-di-dah, the gods have spoken. Some other viewpoints: "The value

By other gods?

>or significance of the results deduced from the axioms was judged, at
>least until severty-five or so years ago, by what they affirmed about the
>physical world." (Kline) "The only consistent point of view...is rather
>the assumption that the source and last raison d'etre of the number
>concept--not only the natural but also the real
>numbers--lie in experience and practical applicability." (Mostowski)
>"If one was willing to accept the sciences, one might as well accept the
>classical system of mathematics." (von Neumann) "How does Peano arrive
>at his own fundamental principles...since after all he cannot prove them
>either? Evidently by analyzing the modes of inference that in the course
>of history came to be recognized as valid and by pointing out that the
>principles are intuitively evident and necessary for science." (Zermelo)

Zermelo wrote this, in "Neuer Beweis fuer die Moeglichkeit einer
Wohlordnung", Math. Ann. 1908; and a few lines below, he spelled out in
detail what he meant by "necessary for science". He said, necessary for
showing

the sums of equivalent disjoint sets are equivalent;
a Dedekind-finite set is finite;
a countable union of countable sets is countable;
R has a basis over Q;
what are the solutions of f(x+y) = f(x) + f(y);

Well now, what a coincidence... Leaving that aside, it is very
clear to what "science" these matters belong. "Necessary for science"
does not refer to the science of Physics, or Chemistry, or... Right?

It refers to Mathematics.

You might say this is not obvious, I suppose. In any case, taking
quotes out of context to try to create false impressions does not always
work. That's obvious.

>Let's give a summary...
>
>You asked for grounds on which AC "might be questionable".
>
>I reply you are the one who should provide the grounds. By Occam's
>Razor, it is the person who is accepting, not the person who is agnostic,
>who should supply the reasons. You make a joke about Occam's Razor. Ha ha.

Hilarious, yes... that you think I need your approval, and that since
you feel I am not entitled to ask the question, I may not ask it.

I did ask it; and others responded, on the subject.

>I present two possibilities for accepting an axiom: (1) it is
>intuitively obvious; (2) it is useful in some way.
>
>You ignore (2)--which, accept my advice, is clearly the way to go--and
>instead you say that the AC is intuitively obvious. Well, not quite, AC
>is "natural and plausible". Now most people wouldn't say that
>intuitively obvious is the same as natural and plausible, but apparently
>that's not a problem.

Idle rhetoric and avoiding the issues do not cut it in Mathematics.

>At the end of the day, you have provided two reasons to accept AC:
>(1) It is natural and plausible;
>(2) Most mathematicians accept it.
>
>(2) is clearly (may I say obviously?) wrong because mathematicians have
>been wrong before. You (or someone else, excuse me if my memory fails
>me) think that doesn't matter, because mathematicians *cannot* be wrong
>in this case, because AC is independant of the other axioms. But that
>isn't the conclusion you need; the conclusion you need is that we should
>accept AC. And so the argument can't be right, because there exists x
>such that most mathematicians have accepted x in the past, yet we should
>not accept x. (e.g. take x = all continuous functions are almost
>everywhere differentiable). You can come back and say your argument is:

Only in your imagination. You have yet to mention _one_ mathematician
who accepted x.

> if most mathematicians accept x and x is independant of the other
>axioms, then we should accept x. But that dosn't hold either; take x =
>Euclid's Parallel Line Postulate, which you yourself admit we should not
>accept if we want to do a certain kind of non-Euclidean geometry.

(2) was, mathematicians accept it _as a natural, plausible, funda-
mental principle_.

>So that leaves (1). Truth to say, I do not find AC natural and
>plausible, but you have set the bar so low, I will grant you the
>possibility that *you* can find it natural and plausible. But that
>wasn't the original point, which is why we might find AC questionable.
>If I can in good faith find AC is not natural and plausible, and you have
>no other justificiation than (1), then AC is questionable.

It's the same issue -- is AC natural and plausible, or is it
questionable to a lesser or greater degree?

I'm sure you are saying in good faith that you don't find it
natural and plausible. But that doesn't make AC questionable, it only
makes it questionable for you. That is why I asked for articulable
reasons. Apparently you object to more than AC, i.e. to infinite sets;
but you didn't take up any of the examples and say something more specific;
only that "you don't find it obvious". Fine, but that conveys no informa-
tion to anybody else.

>In any case, we're arriving full circle and not really advancing. I
>suggest you make one more response (if you want), and then we drop it!

Ilias

[Bill Taylor]

ika...@alumni.caltech.edu (Ilias Kastanas) writes:

|> >Saying "every set is constructible" is NOT required in order to "ensure"
|> >there are no unconstructible sets! Just leave it that you know there aren't
|> >any, and forget about making it an axiom.

|> But Bill, if we decide there are no undefinable or non-constructible

|> sets, that does mean every set is constructible... how can we forget about it?

I'm not asking you to forget about it, I'm asking you to *remember* it!

Just that, in fact. Remember it; don't codify it. For one thing, the
concept of definable sets *cannot* be codified, within the theory itself,
for the usual reasons. It's this inside-outside problem again. From outside
the theory we can see every set is definable, but from inside we can't even
talk about this. It is even possible that (by L-Skolem) the set of all
possible reals is countable, (I'm not saying it *is* so, but that it's at
least conceivable), but from inside we cannot possibly prove or "see" this.
The inside/outside distinction is vital here.

Now - does this apply to the constructible sets, L? Firstly, let's get rid
of one misapprehension that some may hold - the (Godel-)constructible sets, L,
are NOT the same as the definable sets. Not a bit. Neither includes the other,
in fact, AFAICS. Penelope Maddy explored this (and other) matters in her
excellent "definabilism" paper in JSL about 2 years ago.

Firstly, there are a great many constructible sets that are not definable;
all the nasty ones come in once you get above the level of w_1^CK, as I
indicated in my "Richard paradoxes" post. The sets above here cannot be
"defined" in a way clear to meta-mathematical reasoning, because that would
involve actually *doing* a definition using specific ordinal notations above
constructive ordinals, which is per se undoable. But also, there are (?)
definable sets which are not constructible. (I may be wrong here, but I
think this is so - I'm (clearly!) not an expert in this.) The reals, for
instance. R is not in L, is it? L has it's own opinion of what R is, but I
don't think it's the same as ours, is it? So the constructible sets and the
definable ones are quite different, though naturally they overlap considerably.

Now I prefer to speak of the definable ones. These are (at most) the "real"
sets. Fred prefers the constructible sets, but maybe he might switch his
affections the other way, if I have been convincing just now, above. So, there's
no use telling me, Ilias, that L is absolute, must exist etc; that cuts no ice.
Maybe it must exist in ZF, but that doesn't *necessarily* mean it must exist
in ontological space, as ZF is just a convenience from my (and Fred's) POV.
L has no existence, no meaning that I can see, for "real" math.

ZF, and V, are fine, as long as they tell us "universal" things about real
math, (PA and analysis), but not necessarily when they get existential. And
AC is "building non-definability into the system", as I said before. Not good.
Fine for doing (some, not all) surreal math, but misleading when they give
existential results about real math.

|> So we are not bringing in an evil, wellordered hierarchy and imposing it,
|> from outside. It is just that L is naturally wellordered... as "definable"
|> things are wont to be.

L itself is definable, but not all its members are! Those that require
nonconstructive ordinals to refer to, may not be.

|> So if we accept every set is constructible, all this is inevitable.

Aaaarrrgghh - but we do not, not I! Yes, every set is definable, but NOT
constructible, (if I read Maddy right). As I said before, adopting V=L as an
axiom is doing more than it innocently appears - it is *also* saying that there
is a "uniform" or "canonical" way of setting up each set in the L hierarchy.
But it needn't be so.

Let me come at it another way.

Let's look at AC, in the class form, "V can be well-ordered".
(This class form is strictly stronger than AC, but no matter, we can do the same
with any V-level.) This means, there is a *single* function, that universally
chooses from every set.
c: P(V)-0 --> V ; c(x) in x.
(I think P(V) = V, but no matter.)
This means there is now a *named* set of members from every collection of sets.
What a nonsense! ZF, even ZFC itself, tells us that there is *no* definable
choice function, even for as little as P(R). So to arbitrarily bung one in, via
this totally mysterious and essentially undefinable c function, is just crazy.

A folly without warrant.

Does this help make my view comprehensible?

-------------------------------------------------------------------------------
Bill Taylor W.Ta...@math.canterbury.ac.nz
-------------------------------------------------------------------------------

Don't anthropomorphize nature. She really hates that.
-------------------------------------------------------------------------------

[Bill Taylor]

Fred Galvin <gal...@math.ukans.edu> writes:

|> > Forgive me for this observation, Fred, but I think this is a bit ingenuous,
|>
|> Not sure what you mean by "ingenuous". My dictionary says: ingenious;
|> noble, honorable; showing innocent or childlike simplicity and candidness;
|> lacking craft or subtlety;

Aha! We can both be happy then. :-) You can make it noble, and I'll
stick to childlike simplicity!

|> > Saying "every set is constructible" is NOT required in order to "ensure"
|>

|> Say what? Would you mind going over that one more time?

See my other podt, in reply to Ilias.

|> However, we Occamists regard
|> theoretical entities, such as sets, as a necessary evil, to be used as
|> sparingly as possible.

Fred - we are entirely on the same wavelength here!

|> It would be desirable to dispense with sets
|> entirely, since they don't really exist in the physical world.

If I may be permitted, the physical world is almost entirely irrelevant.
Math is the study of things in the "logical" world, maybe Popper-3, maybe less
than that. What I called "ontological space". (I really like that term!)

Also, a great many sets *do* exist in this world. But not *all* those in ZF.
Probably not even all the definable ones; but that's another thread.

|> If we can't
|> get along without sets, then let's have as few of them as we can and still
|> have a smooth-running theory.

You my man, man!

|> Thanks to Goedel, we know that the
|> constructible sets are enough for all practical purposes.

Well actually, not Godel; probably Zermelo. Any low level of V is already
enough. V_2w is often quoted.

|> Therefore, we
|> Occamists only use the constructible sets (and damned few of those, but
|> that's beside the point),

Yes, 100%. But we shouldn't even *refer* to the concept constructible sets,
that's already got a whiff of non-constructibility about it. (That Richard!!)

|> and we see no need to discuss any other kind.

Of course, for real math. But surreal math is fun too!

|> Since constructible sets are the only ones we talk about, we just call
|> them "sets"

Yes, *yes*, YES! Just call them sets, and forget the word "constructible"
altogether! But you can't have AC; see my other post, & the magic function c.

|> Some anti-Occamists, on the other hand, are enamored of sets, can't have
|> enough of them. Some of them, I think, believe that Cantor's Absolute
|> really exists: that God created an immense universe of sets, and it is our
|> religious duty to try to learn as much about it as we can.

Oh come come, Fred. Don't be such a puritan. Or, as you can't help it,
being a physics-mathie, at least be tolerant of the foibles of your more
popish brethren. Ontological space is the mind of god. (You must here imagine
me "crossing" myself with motions in the shape of a Boyes' surface.)

|> The infidel
|> set-theorists just think set theory is fun, and they want to have lots of
|> sets to do surreal math with.

Natch! Fun is fun. It's fun having fun! Especially on someone else's money!

|> It seems to me that, if you're going to admit *some* nonconstructible
|> sets, you should let them *all*.

I admit none as real math; anything (consistent) you like, off and on, as
surreal math. But Fred, I think you may still be confusing constructible
with definable. Apologies if I'm wrong about this.

-------------------------------------------------------------------------------
Bill Taylor W.Ta...@math.canterbury.ac.nz
-------------------------------------------------------------------------------

The chief difference between mathematics and physics is that
in mathematics we have much more direct contact with reality.
-------------------------------------------------------------------------------

[Ilias Kastanas]

In article <5vnr07$22t$1...@cantuc.canterbury.ac.nz>,

Bill Taylor <mat...@math.canterbury.ac.nz> wrote:
>ika...@alumni.caltech.edu (Ilias Kastanas) writes:
>
>|> >Saying "every set is constructible" is NOT required in order to "ensure"
> >there are no unconstructible sets! Just leave it that you know there aren't
>|> >any, and forget about making it an axiom.
>
>|> But Bill, if we decide there are no undefinable or non-constructible
> sets, that does mean every set is constructible... how can we forget about it?

>I'm not asking you to forget about it, I'm asking you to *remember* it!
>
>Just that, in fact. Remember it; don't codify it. For one thing, the
>concept of definable sets *cannot* be codified, within the theory itself,
>for the usual reasons. It's this inside-outside problem again. From outside
>the theory we can see every set is definable, but from inside we can't even
>talk about this. It is even possible that (by L-Skolem) the set of all
>possible reals is countable, (I'm not saying it *is* so, but that it's at
>least conceivable), but from inside we cannot possibly prove or "see" this.
>The inside/outside distinction is vital here.

Well, we are not talking about defining truth; just objects. There
is no a priori obstacle to _that_. Yes, there are L-S phenomena about non-
absolute notions like cardinality; but definitions can be absolute (e.g.
the one of L is). In the case of L we do see from inside L that everything
is constructible, just like we see it from outside as well.

I assumed that "definable", if not the same as "constructible", was
at least contained in it. Maybe it isn't so. Okay then, the question is:
what do you mean by "definable"? When we see, from outside the theory,
that every set is definable... what is it that we "see"?

>Now - does this apply to the constructible sets, L? Firstly, let's get rid
>of one misapprehension that some may hold - the (Godel-)constructible sets, L,
>are NOT the same as the definable sets. Not a bit. Neither includes the other,
>in fact, AFAICS. Penelope Maddy explored this (and other) matters in her
>excellent "definabilism" paper in JSL about 2 years ago.

I didn't see it, unfortunately. Anyway, we are in search of the
notion "definability"...

>Firstly, there are a great many constructible sets that are not definable;
>all the nasty ones come in once you get above the level of w_1^CK, as I
>indicated in my "Richard paradoxes" post. The sets above here cannot be
>"defined" in a way clear to meta-mathematical reasoning, because that would
>involve actually *doing* a definition using specific ordinal notations above
>constructive ordinals, which is per se undoable. But also, there are (?)
>definable sets which are not constructible. (I may be wrong here, but I
>think this is so - I'm (clearly!) not an expert in this.) The reals, for
>instance. R is not in L, is it? L has it's own opinion of what R is, but I
>don't think it's the same as ours, is it? So the constructible sets and the
>definable ones are quite different, though naturally they overlap considerably.

All right, so you intend effectiveness/computability as part of
"definable". It's easy to be misunderstood here; some might accept first-
order definability in set theory, or at least second-order in analysis,
e.g. ExAy Q (Sigma-1-2; x, y real) and the like.

Now since you are objecting to what can only be defined by going
past w_1-CK, you probably wouldn't agree with them. The reals definable up
to w_1-CK are, as you know, the Delta-1-1 (or: HYP) reals, those that are
both Pi-1-1 and Sigma-1-1. And they do have a strong claim to being what
you get if you extend "recursive" as far as possible without going non-con-
structive: they are the same as the reals with arithmetical definitions,
Sigma-0-n, and their transfinite extensions, Sigma-0-a (which is a nontri-
vial theorem, Suslin-Kleene).

Of course there might be other requirements. "Effective" in the
sense HYP is does not also mean "predicative", for one thing.

The reverse, whether something non-constructible can be deemed
"definable", is the crucial question.

The other reals of L appear at various stages after w_1-CK; all of
them have shown up by level w_1 (Goedel's proof of CH). You ask, those
"reals of L" are not "our R", are they? Well, that's independent!... be-
cause V=L _is_ consistent, and under it the answer is yes, that's all of
R; and ~ V=L is also consistent, in the form V=L[z], where z is a
non-constructible real (by forcing).

By the way, MC = "there is a measurable cardinal" does imply ~ V=L
and produces 0#, a non-constructible Delta-1-3 real. But I'll go out on
a limb and say you might not be too sympathetic with MC...

Anyway, all Sigma-1-2 reals are in L, so any non-L real would have
to be at least that bad. Do you see a notion of "definable" that would
cover such a thing? I'm not too optimistic about it...

>Now I prefer to speak of the definable ones. These are (at most) the "real"
>sets. Fred prefers the constructible sets, but maybe he might switch his
>affections the other way, if I have been convincing just now, above. So, there's
>no use telling me, Ilias, that L is absolute, must exist etc; that cuts no ice.
>Maybe it must exist in ZF, but that doesn't *necessarily* mean it must exist
>in ontological space, as ZF is just a convenience from my (and Fred's) POV.
>L has no existence, no meaning that I can see, for "real" math.

I used "absolute" as a technical term; let me try to understand
what you mean here.

The Goedel operations, x-y, {x,y} ... themselves seem hardly con-
troversial; applied to definable objects they yield definable objects.
(Do you agree?!) So when you say L is not "real math" I imagine you refer
to "long t.i.'s" over ordinals. That is, L_(w_1-CK) _is_ "real math",
isn't it?

I'm not too sure in what sense ZF is just a convenience. If it
doesn't codify what we do in Math "legitimately", in what way does it
go overboard? How would you limit it to make it "more realistic"?

I don't quite get what exactly you are proposing as real math,
where only definables exist, Bill! Let us say, is Cantor's |w| < |P(w)|,
|P(w)| < |PP(w)|, ... real? Do we take "all" of P(w)? Is every member
of P(w) definable? I don't know what you intend here.

>ZF, and V, are fine, as long as they tell us "universal" things about real
>math, (PA and analysis), but not necessarily when they get existential. And
>AC is "building non-definability into the system", as I said before. Not good.
>Fine for doing (some, not all) surreal math, but misleading when they give
>existential results about real math.

Yes, universal statements (Pi-1, Ax Q with Q bounded) are downward
absolute: if they hold in a model, they hold in submodels. So ZF is inno-
cuous in that regard; it proves them about V... they hold for us in "less
than V". Hmm... Pi-1 theorems of ZF then are "trustworthy"...

In the classical setting, I should think non-definability is already
there, before AC's building it in. Now what about the "definable" world?
Let's get minimal (!) and take AC_w -2 (countable choice for pairs).
Could you give a short account of it (its failure, I presume) in that world?

>|> So we are not bringing in an evil, wellordered hierarchy and imposing it,
>|> from outside. It is just that L is naturally wellordered... as "definable"
>|> things are wont to be.
>
>L itself is definable, but not all its members are! Those that require
>nonconstructive ordinals to refer to, may not be.
>
>|> So if we accept every set is constructible, all this is inevitable.
>
>Aaaarrrgghh - but we do not, not I! Yes, every set is definable, but NOT
>constructible, (if I read Maddy right). As I said before, adopting V=L as an
>axiom is doing more than it innocently appears - it is *also* saying that there
>is a "uniform" or "canonical" way of setting up each set in the L hierarchy.
>But it needn't be so.

OK Bill, I understand you are not saying V=L.

For the record, the L hierarchy is set up and wellordered the way
it is regardless of whether V=L or not.

>Let me come at it another way.
>
>Let's look at AC, in the class form, "V can be well-ordered".
(This class form is strictly stronger than AC, but no matter, we can do the same
>with any V-level.) This means, there is a *single* function, that universally
>chooses from every set.
> c: P(V)-0 --> V ; c(x) in x.
>(I think P(V) = V, but no matter.)
>This means there is now a *named* set of members from every collection of sets.
>What a nonsense! ZF, even ZFC itself, tells us that there is *no* definable
choice function, even for as little as P(R). So to arbitrarily bung one in, via
>this totally mysterious and essentially undefinable c function, is just crazy.

Just a stab in the dark, but I'm getting the impression that the
c function might not have your full faith and confidence after all...

Eh, yes, it is _consistent_ with ZFC that there be no definable
choice for P(R) (= no definable wellordering of R). It cannot be provable,
though, since L models ZFC.

The issue, however, can be troubling for the "definable" world. It
is typical for any notion of definability to admit a wellordering of its
definitions. So, what happens in it?

>A folly without warrant.
>
>Does this help make my view comprehensible?

Hmm... juxtaposing these two lines could be unwise...?! Anyway
I hope there is some method in my madness!

Ilias

[Matthew P Wiener]

In article <5vnr07$22t$1...@cantuc.canterbury.ac.nz>, mathwft@math (Bill Taylor) writes:

>ZF, and V, are fine, as long as they tell us "universal" things about real
>math, (PA and analysis), but not necessarily when they get existential. And
>AC is "building non-definability into the system", as I said before. Not good.
>Fine for doing (some, not all) surreal math, but misleading when they give
>existential results about real math.

So what do make of Diaphontine equations whose solutions, if any, are codes
for proofs that measurable cardinals are inconsistent with ZF?
--
-Matthew P Wiener (wee...@sagi.wistar.upenn.edu)

[Brian M. Scott]

On Wed, 17 Sep 1997 01:02:43 GMT, sc...@math.csuohio.edu (Brian M.
Scott) wrote:

>On 15 Sep 1997 05:19:06 GMT, mat...@math.canterbury.ac.nz (Bill
>Taylor) wrote:

>>On this general topic, there are some remarkably tricky things proved with AC
>>in ordinary R^3. Things like, "it is possible to partition 3-D space into
>>lines all with different slopes"; or "circles all with nonparallel planes";
>>or other things like so. Alas, my colleague who told me about such matters
>>many years ago can't recall them now!

>>Can anyone else help?

>The first one's not too hard; I haven't thought about the second, but
>I suspect that it's similar.

It is. Enumerate R^3 in type 2^w. At stage k < 2^w, let x(k) be the
first point in the enumeration that isn't in the union of the k
circles C(m), m < k, that have already been chosen. Pick a plane P
containing x(k) and not parallel to any of the C(m), and let L be any
line in P containing x(k). For each m < k, C(m) meets P in at most 2
points; let A be the set of all such points of intersection. For each
y in A there is a unique z in L such that the circle with centre z
passing through x(k) contains y. Since |A| < 2^w, we may choose p in
L different from all such z and take C(k) to be the circle with centre
p containing x(k). Clearly the induction goes through to 2^w and the
C(k), k < 2^w, partition R^3 into pairwise non-parallel circles.
(Note that by exercising just a little care in the choice of the plane
P we can ensure that each plane in R^3 is parallel to one of the
C(k).)

Brian M. Scott

[Bill Taylor]

I knew I shouldn't have got into this thread! It has all the hall-marks of
Theseus' thread through the labyrinth. (But who is the bull?)

Most of the points on either side have been answered already, but always
there seems something more to say!

ika...@alumni.caltech.edu (Ilias Kastanas) writes:
|
|> >Now - does this apply to the constructible sets, L? Firstly, let's get rid
|> >of one misapprehension that some may hold - the (Godel-)constructible sets, L,
|> >are NOT the same as the definable sets. Not a bit. Neither includes the other,
|> >in fact, AFAICS. Penelope Maddy explored this (and other) matters in her
|> >excellent "definabilism" paper in JSL about 2 years ago.
|>
|> I didn't see it, unfortunately.

Oh, pity. Almost required reading for this thread! :) Sorry I can't recall
the date, but it was a big article, the "lead" article of its issue, within
the last 3 years at most, probably 2. Someone reading this thread must have
a pile of JSLs in their office, to tell us?

|> All right, so you intend effectiveness/computability as part of
|> "definable".

Oh no. Not at all. Plenty of non-constructive things are perfectly well
definable! e.g. the set of non-halting machine numbers.

|> Just a stab in the dark, but I'm getting the impression that the
|> c function might not have your full faith and confidence after all...

:-) :-) :-) Keep doing this, Ilias, and I'll be reading your posts
whatever they're about.

|> Eh, yes, it is _consistent_ with ZFC that there be no definable
|> choice for P(R) (= no definable wellordering of R).

Yes; please keep picking my nits, Ilias, or I'll get flea-ridden. As you
must've noticed, when I say such-&-such is so, I often mean is consistent.
Manytimes seems much the same when you relax into definitionalism.

|> It is typical for any notion of definability to admit a wellordering of its
|> definitions.

Weeeeeeell, not really... It's a bit Richardy again. Mustn't confuse theory
and meta-theory, that's at the heart of these problems. Inside and outside!

|> >A folly without warrant.
|> >
|> >Does this help make my view comprehensible?
|> Hmm... juxtaposing these two lines could be unwise...?!

<SNORT!> ;-)

|> Anyway I hope there is some method in my madness!

Well as you know, we're both cat lovers, so there's probably not a lot of hope...

-------------------------------------------------------------------------------
Bill Taylor W.Ta...@math.canterbury.ac.nz
-------------------------------------------------------------------------------

Now that I've given up hope, I feel *much* better!
-------------------------------------------------------------------------------

[Bill Taylor]

wee...@sagi.wistar.upenn.edu (Matthew P Wiener) writes:

|> >ZF, and V, are fine, as long as they tell us "universal" things about real
|> >math, (PA and analysis), but not necessarily when they get existential. And
|> >AC is "building non-definability into the system", as I said before. Not good.
|> >Fine for doing (some, not all) surreal math, but misleading when they give
|> >existential results about real math.
|>

|> So what do make of Diaphontine equations whose solutions, if any, are codes
|> for proofs that measurable cardinals are inconsistent with ZF?

Ah yes, I think Cohen in his CH book made points of this nature. Great and
very real things, Phiodantine equations.

What do I make of it? I'm not sure what kind of answer you want.
What *should* I make of it? What are my options?

-------------------------------------------------------------------------------
Bill Taylor W.Ta...@math.canterbury.ac.nz
-------------------------------------------------------------------------------

Cynicism can be cured - but who cares - it wont help.
-------------------------------------------------------------------------------

[Bill Taylor]

sc...@math.csuohio.edu (Brian M. Scott) writes:

|> It is. Enumerate R^3 in type 2^w. At stage k < 2^w, let x(k) be the

|> first point in the enumeration that isn't in the union of the k circles...
... ...

|> and take C(k) to be the circle with centre p containing x(k).

EXCELLENT!! That's exactly the thing I was looking for, that I dimly recalled.

Great, many thanks to you and the others who provided examples.

Now, as rthe reward for doing one job well is usually being asked to do a
harder one, (C.S.Lewis, "Horse & His Boy), here is a tougher one; but I'm sure
you'll crack it...

Partition R^3 into a collection od circles which are pairwise linked.

-------------------------------------------------------------------------------
Bill Taylor W.Ta...@math.canterbury.ac.nz
-------------------------------------------------------------------------------

...Work'd by wind Pow'r or wat'ry Force.
Or by a circumambulating horse.
George Cockings, 1761.
-------------------------------------------------------------------------------
(Oh well... use Hilbert's horse if you have to...)

[Bill Taylor]

ika...@alumni.caltech.edu (Ilias Kastanas) writes:

|> Well, the issue here is that (as Fred Galvin has also mentioned)
|> if we only "accept" what is definable, then Wellordering holds -- in fact
|> there is a global wellordering of the universe.

No and no. As explained before.

|> This is certainly so if
|> "definable" means first-order (or anything less); we end up with the con-
|> structible sets (L).

No, as mentioned.

|> But we have already indulged in the "surreal" in accepting the set
|> of all reals; surely only some of them can be definable.

No, they are all definable. Any individual real (that can be mentioned (in the
theory)) is automatically per se definable. In ont. space there are no others.

|> stock and barrel? Note also that there can be sets of reals that are
|> definable without even a single real in them being definable.

Indeed, I think I alluded to this point already.

Extra bit: lost the excerpt, sorry:- But in this post or the other one you
asked - "What exactly do you mean by definable?" I think it's a standard
usage; see e.g. Levy "Basic Set Theory" - there's a nice section on it.

|> what I'm saying is that once we accept infinite
|> sets the way we have, AC follows from the same intuition, not an extension.

And I'm saying it doesn't. Impasse. Banana skins at 40 paces!

|> We collect "all subsets of w" to form R, without giving an effective
|> procedure that will produce them, one after another;

Sure, I don't mind that, as long as it is a definable procedure. That same
confusion... constructive(effective) =/= definable =/= Godel-constructible.

|> Hmm... like, every continuous positive function on [0,1] has a
|> positive l.u.b., but we may not assert it occurs at a definite x...

Hell - no problem! Just define the sequence of half-intervals that contain the
sup(f); (left-hand one if both). Totally definable; totally non-constructive.

|> But the view that AC_w is intuitively "more" justifiable, maybe to the
|> point of granting AC_w while denying AC, is something I don't get.

Fair point. Strictly - it isn't tenable in *all* situations - but it requires
extraordinary effort to find counterexamples, (e.g. using w_1^CK); so using
it is harmless, because standard math is never going to see these. But AC
has counterexamples all over the shop: no WO-of-R, f'rinstance.

|> Okay, Occam's razor cuts both ways...

It certainly does! That's why I'm not so very wild about it. A classic case
arises in the endless many-worlds discussions of QM on sci.physics. One side
invokes Occy to get rid of the umptagazillion unseen worlds, (but needs to
have a new principle - some sort of reduction); & t'other side invokes Billy-O
to eliminate unneeded reduction, and is then saddled with extreme ontology!

It seems Occam for theory and Occam for existence are contravariant!

|> A stray note:

I caught it!!

|> AC actually makes all sets be of a rather concrete kind!

Indeed so! But as I said to Fred, it does more - it insists on this hidden
"canonical form" for them, which is a big no-no.

|> > There are no objects such that there are no such objects.

|> Eh... such a statement is well-liked by people who would like
|> this sort of statement.

Alice: I haven't had *any* yet, so I can hardly have *more*.
Hatter: On the contrary! It's very easy to have *more* than nothing...

|> Yes, yes, a good illustration of the evil nature of AC, right?!

"Luke, Luke - you don't know the power of the dark side of the axiom!"

-------------------------------------------------------------------------------
Bill Taylor W.Ta...@math.canterbury.ac.nz
-------------------------------------------------------------------------------

I think math is important, but I don't think it's as important as I think it is.
-------------------------------------------------------------------------------

[Matthew P Wiener]

In article <5vq58r$pk9$2...@cantuc.canterbury.ac.nz>, mathwft@math (Bill Taylor) writes:
>
>wee...@sagi.wistar.upenn.edu (Matthew P Wiener) writes:

>|> >ZF, and V, are fine, as long as they tell us "universal" things
>|> >about real math, (PA and analysis), but not necessarily when they
>|> >get existential. And AC is "building non-definability into the
>|> >system", as I said before. Not good. Fine for doing (some, not
>|> >all) surreal math, but misleading when they give existential
>|> >results about real math.

>|> So what do make of Diaphontine equations whose solutions, if any,

>|> are codes for proofs that measurable cardinals are inconsistent
>|> with ZF?

>What do I make of it? I'm not sure what kind of answer you want.

>What *should* I make of it? What are my options?

Your options are to make something of it. Your view suggests there is a
clean break between Real Math and Surreal Math, and that you can happily
camp out in the Real Math territory and not think about the question.
There isn't such a clean break.

You can, if you wish, not think about the question, but if so, it is hard
to take your foundational philosophizing as serious or even sincere.

[Daryl McCullough]

Bill, I still don't understand the intuition behind your claim that the
only sets that exist are the definable ones. And I also don't understand
why you say why this claim implies that the axiom of choice is false.

Oh, actually, I guess it depends on whether you mean that the only sets
that exist are the *hereditarily* definable ones. (That is, sets that
consist of only definable sets.) It seems to me that there can't
possibly be more than a countable number of hereditarily definable sets
(if you don't allow set parameters in your definitions). So the only
axiom of choice that you need is countable choice, which is much weaker.

But maybe you don't mean hereditarily definable, but simply definable.
The set of all reals is definable in set theory, so this set exists and
is uncountable (in contrast with the set of all definable reals).
However, since there are only countably many definable reals, and those
are the only reals that exist, in your ontology, it would seem to follow
that (in some sense) *most* elements of the set of all reals don't
exist! I just don't see the intuition behind that.

I can understand much better the attitude of Bishop, who didn't *deny*
that unconstructible objects exist (since for him, denying existence of
something requires a proof of nonexistence, which was impossible).
Instead, he just tried to do as much mathematics as he could without
relying on such objects.

Daryl McCullough
CoGenTex, Inc.
Ithaca, NY

[Jim Hunter]

Bill Taylor wrote:

> |> Okay, Occam's razor cuts both ways...
>

> It certainly does! That's why I'm not so very wild about it. A classic case
> arises in the endless many-worlds discussions of QM on sci.physics. One side
> invokes Occy to get rid of the umptagazillion unseen worlds, (but needs to
> have a new principle - some sort of reduction); & t'other side invokes Billy-O
> to eliminate unneeded reduction, and is then saddled with extreme ontology!
>
> It seems Occam for theory and Occam for existence are contravariant!
>

Sorry about butting in on your tread, but MW in flawed because it
violates the most sacred conversation law of all. That at the beginning
of a thought and at the end of a thought, we preserve the number of
minds
we count at one.

---
Jim

[Ilias Kastanas]

In article <5vq71c$rej$1...@cantuc.canterbury.ac.nz>,

Bill Taylor <mat...@math.canterbury.ac.nz> wrote:
>
>ika...@alumni.caltech.edu (Ilias Kastanas) writes:
>
>|> This is certainly so if
>|> "definable" means first-order (or anything less); we end up with the con-
>|> structible sets (L).
>

>No, as mentioned.

Well, it's interesting to have a notion of definability that
produces non-constructible objects too... I would like to see an example
or an indication!

>|> But we have already indulged in the "surreal" in accepting the set
>|> of all reals; surely only some of them can be definable.
>

>No, they are all definable. Any individual real (that can be mentioned (in the
>theory)) is automatically per se definable. In ont. space there are no others.
>
>

>Extra bit: lost the excerpt, sorry:- But in this post or the other one you
>asked - "What exactly do you mean by definable?" I think it's a standard
>usage; see e.g. Levy "Basic Set Theory" - there's a nice section on it.
>
>

>|> what I'm saying is that once we accept infinite
>|> sets the way we have, AC follows from the same intuition, not an extension.
>

>And I'm saying it doesn't. Impasse. Banana skins at 40 paces!

Not necessarily! The above comes from the "stages" motivation for
set theory (a set can be formed at stage S out of anything already formed
in stages prior to S... etc; no "minimality"); in the standard approach,
AC is just as immediate a consequence as Union or Powerset. I understand
you are saying "not so" under the "definables!" approach. Let me try to
see how this works:

>|> We collect "all subsets of w" to form R, without giving an effective
>|> procedure that will produce them, one after another;
>

>Sure, I don't mind that, as long as it is a definable procedure. That same
>confusion... constructive(effective) =/= definable =/= Godel-constructible.

OK, x is a real if Ay y in x -> y in w; I take it, this is
"definable" and hence we admit all such x to form R.

Now, f is a tuple if Ay y in J -> f(y) in x(y); isn't this simi-
lar, and ground for admitting f and forming the product of x(y), (y in J)?
No "rule" for _what_ element of x(y)... just like there was no rule for
which elements of w will be in a real. An f seems no less "definable"
than a real! What is going on?!

>|> But the view that AC_w is intuitively "more" justifiable, maybe to the
>|> point of granting AC_w while denying AC, is something I don't get.
>

>Fair point. Strictly - it isn't tenable in *all* situations - but it requires
>extraordinary effort to find counterexamples, (e.g. using w_1^CK); so using
>it is harmless, because standard math is never going to see these. But AC
>has counterexamples all over the shop: no WO-of-R, f'rinstance.

How about a hint as to how no-WO-of-R is obtained?

By the way, Bill, you asked once about the relative strength
of (below-AC_w material, natch!... like):

(1) Every countable union of countable sets is countable
(2) Every infinite set has a countable subset (= no Dedekind cardinals)

Pulling together what I posted here and there, we have:

(1) does not imply (2)

There is a model (by forcing) where an infinite set of reals vio-
lates (2). All the following hold in the model, and hence fail to imply (2):

Any family has an f with f(x) _subset_ (non0, proper) of x.
(Hence) every set has a linear ordering.
Choice for any family of countable sets. In fact, more: it is enough
if each set is equivalent to some aleph or other.
w_1 is regular (not countable union of countable ordinals)
Finally, (1). It follows from above.

(2) does not imply (1)

By another model, of (2) and ~(1). (2) implies countable unions
of finite sets are countable... but not (1).

Of course neither (1) nor (2) implies AC_w.

These are the formal results. Now, what happens in the "definable"
approach? I guess you said AC_w doesn't always hold; what about (1) and (2)?
Once again, in spite of independence, such a split seems counterintuitive!

Ilias

[kem...@de.ibm.com]

In <5vigga$il7$1...@cantuc.canterbury.ac.nz>, mat...@math.canterbury.ac.nz (Bill Taylor) writes:

>ika...@alumni.caltech.edu (Ilias Kastanas) writes:
>
>
>|> You and others have such against AC, and I would like to
>|> hear what they are. There have been discussions in the past; what is it
>|> like in the present?
>
>
>Well, my motivation is simple enough. It is merely that the only sets
>which "really exist" are definable ones; (and quite possibly not even all
>of those! But that's a whole nother thread.)
>

O.k. If something "really exists" it's describable/definable in some kind of terms.
But how do you conclude that something really exists ? By picking it out of
what ? Why, for example, should the definite points of occurence of a
photon be restricted to a finite set of space-time locations ?

>
>It seems this was Lebesgue's (& Borel's) idea also; but they lived back before
>such concepts could be neatly codified, (as e.g. in my w_1^CK post, though
>others have done better).
>
>

>So all those lovely choice-necessitous things - Hamel bases, nonmeasurable
>sets, Banach-T partitions, unprincipled infiltrators, etc etc, which ARE
>lovely things, no question, simply "aren't there" (in ontological space!)
>So results about them, and many results using them, are more like "what it
>would be like if these (impossible) things *did* exist". A kind of fantasy
>math. What was Weemba's term for it...? Oh yes, "SURREAL math". Nice.
>

Not that this is about AC but wasn't the Lorentz transformation in ontological
space before the related physical experiments?

>
>Not that I've anything against it at all. It's all legitimate math, math
>doesn't *have* to be fully "real", as we know. I'm all for Hilbert and
>the pure math toast! But still, it's irritating to see surreal math
>passed off as real math, as is automatically assumed in today's texts. Well,
>perhaps not quite automatically... was it Kreisel who said, "AC is unique
>in its ability to trouble the conscience of the ordinary mathematician!"
>

>AC_w being used in 19th-century style math is not nearly so heinous; most
>such cases (of existence proofs), can easily be made constructive for any
>particular example. It's very hard to drum up *really* nonconstructive
>things on the real line; the early intuitionists notwithstanding. But no

>doubt it could be done, with extraordinary effort; it just won't happen in
>bridge-building math. And anyway, there's always the conservative extension
>meta-theorem to fall back on!
>

>So.
>That's about it. It could be said to be a case of ontological minimality,
>to put a pompous label on it, or Occam's razor, to give it a patina of
>scholastic respectability. But it just amounts to noticing that non-definable
>things "just don't exist", however consistent they may be.
>

Good point. Ontological minimality -- isn't this exactly the question on
minimal axiom systems? But on which purposes? I guess just to bring up
a consistent frame for (mathematical) modelling. If there is nothing fiting to
an axiom system ,well, then it may be useless.
But again : Lorentz transformation or the finite (non-euclidean ) geometry
that is (nowadays) highly related to error correcting codes.
Both selected out of which ontological actuallity ?
Don't you think that definability often depends on the available notation tools?

regards

>--------

>
>On this general topic, there are some remarkably tricky things proved with AC
>in ordinary R^3. Things like, "it is possible to partition 3-D space into
>lines all with different slopes"; or "circles all with nonparallel planes";
>or other things like so. Alas, my colleague who told me about such matters
>many years ago can't recall them now!
>
>Can anyone else help?
>

>-------------------------------------------------------------------------------
> Bill Taylor W.Ta...@math.canterbury.ac.nz
>-------------------------------------------------------------------------------

> There are no objects such that there are no such objects.

>-------------------------------------------------------------------------------

[Bill Taylor]

wee...@sagi.wistar.upenn.edu (Matthew P Wiener) writes:

|> You can, if you wish, not think about the question, but if so, it is hard
|> to take your foundational philosophizing as serious or even sincere.

Oh dear! This sounds like bad news for me. Mind you, coming from someone
who has persistently refused to answer queries of mine about Penrose/Lucas
evolution stuff... and in spite of three (count them: 3) *very* specific
requests last year, for a very specific type of post, all ignored...

I recall someone using the term "intellectually lazy". Hmmmmm....

|> Your view suggests there is a
|> clean break between Real Math and Surreal Math,

Yes, I think a division is discernible.

|> and that you can happily camp out in the Real Math territory

I'm happy either way. We don't have to set ourselves in concrete.

|> and not think about the question.

Good heavens! What have we all been doing all this time!

|> There isn't such a clean break.

At the moment, it's not completely clear. But that was the state of
orthodox v constructive math, for many decades. It slowly became clearer,
until at last, with the work of Bishop and Bridges it seems fairly clear what
the essential didfferences are. It may be that the same will happen here.

|> >|> So what do make of Diaphontine equations whose solutions, if any,
|> >|> are codes for proofs that measurable cardinals are inconsistent
|> >|> with ZF?

You'll have to be more explicit. Proofs in what theory?

-------------------------------------------------------------------------------
Bill Taylor W.Ta...@math.canterbury.ac.nz
-------------------------------------------------------------------------------

[Bill Taylor]

Daryl McCullough <da...@cogentex.com> writes:

|> Bill, I still don't understand the intuition behind your claim that the
|> only sets that exist are the definable ones.

I think there is a major problem of not being on the same wavelength here.

The situation is a little like the past debates between constructivists and
orthodox mathematicans. Constructivists used to say, "we're not trying to
*restrict* anything, we're just going with what is *really there*, that is,
constructive. But they had insuperable problems with bivalence, the
status of truth before discovery, the merely socio/historical nature of
assertablitiy, and so forth. Ultimately their view was not convincing,
though they did valuable service (along with Turing et al) in clarifying
the nature and extent of the constructive notion, as a subset of math.

But ultimately, most people agree - a set like "the set of numbers of
non-halting machines" simply IS. Some numbers are in, and some are out,
whether we fallible humans know which or not.

Now my "definitionalist" viewpoint is that *any* set that can be spoken of,
(like the above), probably IS, in some sense (there is room for debate here);
but certainly that any set which CANNOT be spoken of, IS NOT. It is silly
even to speak of such, as none can ever be specifically mentioned.

This view will be incomprehensible to someone who is not on the wavelength -
they will always feel I'm "restricting" sets in some way. But no - I'm
saying - any individual set you can talk about (consistently) is fine - talk
about it. But don't expect there to be any others. Any others IN THE
META-THEORY, that is, the distinction is crucial. Nothing has been restricted,
in the way that constructivists restrict.

|> And I also don't understand
|> why you say why this claim implies that the axiom of choice is false.

Well it seems "intuitively obvious" on this view. There is (e.g.) no well
ordering of R. No set that does this, IS. If there IS such a set, it should
be individually discernable, referable to; but we know this is impossible.
(See Levy, section on definable sets.) (Yes I know - there "may be" such
referable-to sets, so long as it be permanently undetectable that they do what
it's claimed that they might (i.e. wellorder R), but on the definitionalst
view this amounts to the same thing.)

|> Oh, actually, I guess it depends on whether you mean that the only sets
|> that exist are the *hereditarily* definable ones.

An interesting notion of course; but not a vital one I think.

|> (if you don't allow set parameters in your definitions). So the only
|> axiom of choice that you need is countable choice,

No, I think there are still two confusions here. (i) the confusion between
constructive and definable (which I have dealt with several times); (ii) the
confusion between theory and meta-theory (ditto). On the latter, recall
that the whole universe of actually existing sets MAY be even countable,
(it may not too, there are subtle points debatable here, also mentioned before);
but that is COUNTABLE SEEN FROM THE META_THEORY. Internally, in the theory,
of course there are many uncountable sets. So AC-w is not really relevant.

|> However, since there are only countably many definable reals, and those
|> are the only reals that exist, in your ontology, it would seem to follow

|> that (in some sense) *most* elements of the set of all reals don't exist!
, ,
Oh come - anything that exists, exists! Nothing that *doesn't exist*,
can be a memeber of *anything*.

A natural problem! There may (appear to) be only countably many reals, seen
from outside, but there always appear to be uncountably many from the inside.

|> I can understand much better the attitude of Bishop, who didn't *deny*
|> that unconstructible objects exist

I think he did. But he did much worse. He even said things could be false
at one time, and true later on! (Strictly speaking, "not true" earlier and
"true" later, but the ortho-mathie always hear that the way I said it first.)

Cheers,

-------------------------------------------------------------------------------
Bill Taylor W.Ta...@math.canterbury.ac.nz
-------------------------------------------------------------------------------

The intuitionist confuses knowledge with truth;
The constructivist confuses ignorance with impossibility.
-------------------------------------------------------------------------------

[kem...@de.ibm.com]

In <5vli0i$a...@gap.cco.caltech.edu>, ika...@alumni.caltech.edu (Ilias Kastanas) writes:
>In article <5vigga$il7$1...@cantuc.canterbury.ac.nz>,
>Bill Taylor <mat...@math.canterbury.ac.nz> wrote:
>>ika...@alumni.caltech.edu (Ilias Kastanas) writes:
>>
>>|> You and others have such against AC, and I would like to
>>|> hear what they are. There have been discussions in the past; what is it
>>|> like in the present?
>
>>Well, my motivation is simple enough. It is merely that the only sets
>>which "really exist" are definable ones; (and quite possibly not even all
>>of those! But that's a whole nother thread.)
>

> Well, the issue here is that (as Fred Galvin has also mentioned)
>if we only "accept" what is definable, then Wellordering holds -- in fact
>there is a global wellordering of the universe. This is certainly so if
>"definable" means first-order (or anything less); we end up with the con-
>structible sets (L).
>

[snip ]

> By the way, remember the "shotgun marriage" of Wellordering (qua ORD,
>and transfinite induction over it) and Powerset? It's amusing to note that
>Powerset implies arbitrarily large cardinals, and hence ORD and class-length
>t.i.'s; conversely, t.i. over ORD builds L via "definable" Powerset... and
>ends up implying Powerset, since the latter holds formally in L!
>
>>So all those lovely choice-necessitous things - Hamel bases, nonmeasurable
>>sets, Banach-T partitions, unprincipled infiltrators, etc etc, which ARE
>>lovely things, no question, simply "aren't there" (in ontological space!)
>>So results about them, and many results using them, are more like "what it
>>would be like if these (impossible) things *did* exist". A kind of fantasy
>>math. What was Weemba's term for it...? Oh yes, "SURREAL math". Nice.
>
> But we have already indulged in the "surreal" in accepting the set
>of all reals; surely only some of them can be definable. And every w-sequence
>of reals is coded by a real, so we have those too. Isn't it "too late" to
>insist that when it comes to sets of reals, or functions to/from R, P(R), ...
>only "definable" ones can be there? After having admitted R, P(R),... lock,
>stock and barrel? Note also that there can be sets of reals that are
>definable without even a single real in them being definable.
>

[snip]

> It is informative to be able to produce a choice function explicitly
>rather than by appeal to AC; what I'm saying is that once we accept infinite
>sets the way we have, AC follows from the same intuition, not an extension.
>We collect "all subsets of w" to form R, without giving an effective
>procedure that will produce them, one after another; we take any one that is
>there. So if some real is there in each A of a family F, it seems we ought
>to be able to do the same, without a specifying procedure. It's not formally
>provable, of course, but it is based on the same idea.
>
> One can work in ZF, or in various ZF + X - Y; what I'm saying is
>that ZF + AC has the same "surreality" as ZF, from an intuitive viewpoint.
>
>
>>AC_w being used in 19th-century style math is not nearly so heinous; most
>>such cases (of existence proofs), can easily be made constructive for any
>>particular example. It's very hard to drum up *really* nonconstructive
>>things on the real line; the early intuitionists notwithstanding.

[snip]

> And offhand, partition-into-lines looks like a transfinite induction
>similar to, uh, those used for this kind of thing. Suppose lines L_0, ...
>L_a are pairwise non-intersecting (at infinity(!) or otherwise), for an
>ordinal a < c = |R|. Let Q be the least (in the wellordering of R^3) point
>not on any L_i. There is a plane P through Q distinct from the plane through
>Q and L_i, for all i <= a (the least in the wellordering of planes, natch).
>Each L_i intersects P at one point F_i (at infinity or not); let L be a line
>on P through Q, distinct from the a lines QF_i. Define L_a+1 to be L.
>(Similarly if the original lines are L_i, i < a, a limit; L becomes L_a).
>
> For less than c -many L_i, there will always be a Q (easy... though
>"m(L_i) = 0, hence m(union) = 0" is not necessarily correct). But we need
>to be sure L_i, i < c does exhaust R^3 ! That's easy, too; as long as
>R^3 is wellordered in the minimal type c... since the first j points are in
>the union of the first j L_i 's.
> Yes, yes, a good illustration of the evil nature of AC, right?!
>
>
> Ilias

To me it seems that there are different possibilities ( underlying ZF):
1.) Don't use AC ( or even weaker forms e.g. AC_w) as well as no ~AC (not AC).
Thereby (for me) the question arises :
down to which level/stage of incompletness is it in accordance to
an ontological point of view ?
2.) Use AC ( ZF + AC) and you will get all its implications. As far as there is no
contradiction to 'reality' it is an appropriate model.
3.) Use ZF + ~AC. Derive all its wonderfull theorems. If there's no contradiction
to 'reality' it's fine. If (ZF +AC) as well as (ZF+~AC) are not contradicting
'reality' compare their results.
4.) May be there are 'other types of AC' , like there are different
'parallel axioms' , that although might give nice models.

regards
st.l.

[Planar]

>From: mat...@math.canterbury.ac.nz (Bill Taylor)

>
>But ultimately, most people agree - a set like "the set of numbers of
>non-halting machines" simply IS. Some numbers are in, and some are out,
>whether we fallible humans know which or not.
>
>Now my "definitionalist" viewpoint is that *any* set that can be spoken of,
>(like the above), probably IS, in some sense (there is room for debate here);
>but certainly that any set which CANNOT be spoken of, IS NOT. It is silly
>even to speak of such, as none can ever be specifically mentioned.

But isn't this a bit of circular reasoning ? How do you define "can
be spoken of", independently of the axioms ? The goal of the ZF
axioms is to define what sets exists, and can be spoken of, isn't it ?

As far as I can tell, your line of reasoning could be used to deny the
existence of N itself, because it cannot be spoken of, unless you
admit the axiom of infinity.

-- Planar

[kem...@de.ibm.com]

In <5vmtea$aai$1...@belzebul.imaginet.fr>, A Boucher <abou...@imaginet.fr> writes:
>In article <5vlqhm$d...@gap.cco.caltech.edu> Ilias Kastanas,
>ika...@alumni.caltech.edu writes:
>
>>Subject: Re: WEAK CHOICE - another query.
>>From: Ilias Kastanas, ika...@alumni.caltech.edu
>>Date: 16 Sep 1997 11:28:54 GMT
>>>In article <5vk43o$mt9$1...@belzebul.imaginet.fr>,
>>A Boucher <abou...@imaginet.fr> wrote:
>>>In article <5vdd3s$j...@gap.cco.caltech.edu> Ilias Kastanas,
>>>ika...@alumni.caltech.edu writes:
>>>>

[snip]

>>>>>> OK; (0,1) is partitioned into sets E0, E1, E2, ... ; claim: there is
>>>>>>a sequence x0, x1, x2,... (xk in Ek, for k in N); it needs AC_w, so you do not
>>>>>>accept this claim. Any specific reasons you can mention?
>>>>>>
>>>>>> A0, A1, ... are countable sets. Claim: their union (of Ak, k in N)
>>>>>>is countable; it follows from AC_w. Do you reject it? Any reasons?
>>>>>>
>>>>>> Same, with each of A0, A1... having two elements. Now what?
>>>>>>
>>>>>> E is infinite. Claim: there is a sequence x0, x1, ... of distinct
>>>>>>elements of E. (Follows from AC_w; weaker than it). What about this?
>>>>>>

[snip]

>Let's give a summary...
>
>You asked for grounds on which AC "might be questionable".
>
>I reply you are the one who should provide the grounds. By Occam's
>Razor, it is the person who is accepting, not the person who is agnostic,
>who should supply the reasons. You make a joke about Occam's Razor. Ha
>ha.
>

BTW, is it possible to give a brief summary of the founding axiomatic frame
for concluding Occam's Razor ?

>I present two possibilities for accepting an axiom: (1) it is
>intuitively obvious; (2) it is useful in some way.
>

>At the end of the day, you have provided two reasons to accept AC:
>(1) It is natural and plausible;
>(2) Most mathematicians accept it.
>
>(2) is clearly (may I say obviously?) wrong because mathematicians have
>been wrong before. You (or someone else, excuse me if my memory fails
>me) think that doesn't matter, because mathematicians *cannot* be wrong
>in this case, because AC is independant of the other axioms. But that
>isn't the conclusion you need; the conclusion you need is that we should
>accept AC. And so the argument can't be right, because there exists x
>such that most mathematicians have accepted x in the past, yet we should
>not accept x. (e.g. take x = all continuous functions are almost
>everywhere differentiable). You can come back and say your argument is:
> if most mathematicians accept x and x is independant of the other
>axioms, then we should accept x. But that dosn't hold either; take x =
>Euclid's Parallel Line Postulate, which you yourself admit we should not
>accept if we want to do a certain kind of non-Euclidean geometry.
>

Can you please explain what are the differences between a (proved) theorem
and an axiom in your opinion ?

>So that leaves (1). Truth to say, I do not find AC natural and
>plausible, but you have set the bar so low, I will grant you the
>possibility that *you* can find it natural and plausible. But that
>wasn't the original point, which is why we might find AC questionable.
>If I can in good faith find AC is not natural and plausible, and you have
>no other justificiation than (1), then AC is questionable.
>

And if it's not natural and plausible only but even intuitively obvious
it is , isn't it ?

regards

[Matthew P Wiener]

In article <5vt2h5$9cm$1...@cantuc.canterbury.ac.nz>, mathwft@math (Bill Taylor) writes:
>wee...@sagi.wistar.upenn.edu (Matthew P Wiener) writes:

>|> You can, if you wish, not think about the question, but if so, it is hard
>|> to take your foundational philosophizing as serious or even sincere.

>Oh dear! This sounds like bad news for me. Mind you, coming from someone
>who has persistently refused to answer queries of mine about Penrose/Lucas
>evolution stuff... and in spite of three (count them: 3) *very* specific
>requests last year, for a very specific type of post, all ignored...

Oh stuff it. I've answered them repeatedly. Just not the time you popped in.

>|> >|> So what do make of Diaphontine equations whose solutions, if any,
>|> >|> are codes for proofs that measurable cardinals are inconsistent
>|> >|> with ZF?

>You'll have to be more explicit. Proofs in what theory?

The one explicitly mentioned in my question.

[HS Brandsma]

Brian M. Scott (sc...@math.csuohio.edu) wrote:
:
: On 15 Sep 1997 05:19:06 GMT, mat...@math.canterbury.ac.nz (Bill
: Taylor) wrote:

:
: >On this general topic, there are some remarkably tricky things proved with AC

: >in ordinary R^3. Things like, "it is possible to partition 3-D space into
: >lines all with different slopes"; or "circles all with nonparallel planes";
: >or other things like so. Alas, my colleague who told me about such matters
: >many years ago can't recall them now!
: >
: >Can anyone else help?

:
: The first one's not too hard; I haven't thought about the second, but
: I suspect that it's similar. Well-order R^3 in type 2^w. At stage k
: you want to choose a line L(k) though x(k) that has a different slope

: from and is disjoint from each L(m), m < k. (This is assuming that

: x(k) isn't already in the union of the L(m) already chosen; in that

: case there's nothing to do at this stage.)
:
: Each L(m), m < k, determines and is contained in a plane through x(k).
: Since there are fewer than 2^w lines L(m), there is a plane, P, such
: that (1) P contains x(k), and (2) for each m < k, P is not parallel to
: (and therefore in particular doesn't contain) L(m). (For instance,
: let S be the unit sphere - not ball - about x(k). For each m < k let
: A(m) be the 2-point intersection of S with the line through x(k)

: parallel to L(m). Now pick any point y on S that's not in the union
: of the A(m), and let P contain x(k) and y.) Each L(m) meets P in a

: single point. There are fewer than 2^w such points, so there's a line
: in P containing x(k) that misses all of the L(m), m < k; choose such a
: line, and let it be L(k).
:
: Brian M. Scott

Well, actually this doesn't need choice at all. Discussing this problem with
one of our geometers (finite geometry), we came up with the following solution:
R^3 sits naturally in the four-dimensional projective plane (as (x,y,z,1)).
Four dimensional projective plane over R can be considered a 2-dimensional
projective plane over C. This has a trivial partition in lines. Now translate
back to R^3.

This yields the following lines (in parametric notation)
Pick two points u and v, not both 0.
t(0,0,1)
(0,1/u,1/u) + t(1,-v/u, u+(v^2/u)) if u<>0.
(1/v,0,0) + t(0,1,-v) if u=0.

Nice heh!

Henno Brandsma.

(and now circles. ?moebius inversion maybe?)

[Daryl McCullough]

mat...@math.canterbury.ac.nz (Bill Taylor) says...

>Now my "definitionalist" viewpoint is that *any* set that can be spoken of,
>(like the above), probably IS, in some sense (there is room for debate here);
>but certainly that any set which CANNOT be spoken of, IS NOT. It is silly
>even to speak of such, as none can ever be specifically mentioned.

I don't quite know what you mean by that. If you take ZF and add a new
function symbol F and add the axiom

Forall x, x nonempty -> F(x) is an element of x

Then suddenly, voila! Every choice set has a name, so we can speak of
it.

Maybe you would object because F(x) is not uniquely determined by the
axioms. However, that's what I don't understand. Why should every
mathematical object have a name? There are uncountably many reals, for
example, and only countably many names for reals.

>This view will be incomprehensible to someone who is not on the wavelength -
>they will always feel I'm "restricting" sets in some way. But no - I'm
>saying - any individual set you can talk about (consistently) is fine - talk
>about it.

That's easy enough to arrange, by introducing the symbol F described above.

>> And I also don't understand why you say why this claim implies
>> that the axiom of choice is false.
>
>Well it seems "intuitively obvious" on this view.

Not to me.

>There is (e.g.) no well ordering of R.

If by R you mean the definable reals, then there is a well-ordering:
lexicographically order them by their definitions.

>No set that does this, IS. If there IS such a set, it should
>be individually discernable, referable to;

With the F above, you can refer to it.

>> Oh, actually, I guess it depends on whether you mean that the only sets
>> that exist are the *hereditarily* definable ones.
>
>An interesting notion of course; but not a vital one I think.
>
>> (if you don't allow set parameters in your definitions). So the only
>> axiom of choice that you need is countable choice,
>
>No, I think there are still two confusions here. (i) the confusion
>between constructive and definable (which I have dealt with several
>times);

I don't think so. Here's the definition of definable that I'm using:
A class S is definable if there is a formula Phi(x) in ZFC such that
S = { x | Phi(x) }. If S happens to be a set, then it is a definable
set.

>(ii) the confusion between theory and meta-theory (ditto).
>On the latter, recall that the whole universe of actually
>existing sets

We have a language problem right there. What does "actually
existing" mean for sets?

>MAY be even countable, (it may not too, there are subtle
>points debatable here, also mentioned before); but that
>is COUNTABLE SEEN FROM THE META_THEORY.

I don't understand why.

>> However, since there are only countably many definable reals, and those
>> are the only reals that exist, in your ontology, it would seem to follow

>> that (in some sense) *most* elements of the set of all reals don't exist!

>
>Oh come - anything that exists, exists! Nothing that *doesn't exist*,
>can be a memeber of *anything*.

I'm joking, of course. My point is that we have a strange mismatch between
our intuitions about the reals and your definability thesis. There are only
countably many definable reals, while there are uncountably many reals, so
how is it possible that only the definable ones exist?

>A natural problem! There may (appear to) be only countably many reals, seen
>from outside, but there always appear to be uncountably many from the inside.

I would say it this way: it is obviously true that there are uncountably
many reals. (Cantor proved it years ago.) However, since we can only
refer to countably many of them, we are not capable of distinguishing
between the *real* reals and a sufficiently rich countable collection of
reals.

>> I can understand much better the attitude of Bishop, who didn't *deny*
>> that unconstructible objects exist
>
>I think he did. But he did much worse. He even said things could be false
>at one time, and true later on!

No, I think you are confusing Bishop with Brouwer. Bishop was always
careful to only make claims that were classically valid, as well.
He never claimed that nonconstructible things don't exist.

I don't think that, properly understood, there is anything *restrictive*
about constructive mathematics. It is a *richer* mathematics, making
more distinctions than ordinary mathematicians make---a lot like
distinguishing between theorems that use choice and those that don't! 8^)

>-------------------------------------------------------------------------------
> Bill Taylor W.Ta...@math.canterbury.ac.nz
>-------------------------------------------------------------------------------

> The intuitionist confuses knowledge with truth;
> The constructivist confuses ignorance with impossibility.
>-------------------------------------------------------------------------------

My own opinion is that the attitude you attribute to constructivists should
properly be attributed to definabilists. 8^)

Not that I'm anything in particular. I don't think that the question of
the existence of mathematical objects makes any sense whatsoever. What
is interesting to me is whether a mathematical theory, as a whole, is
interesting and useful.

[Fred Galvin]

On 17 Sep 1997, Bill Taylor wrote:

> Fred Galvin <gal...@math.ukans.edu> writes:
>
> |> However, we Occamists regard
> |> theoretical entities, such as sets, as a necessary evil, to be used as
> |> sparingly as possible.
>
> Fred - we are entirely on the same wavelength here!

I doubt it. Sets are convenient fictions, like epicycles in planetary
theory. If the theory works better with one epicycle more or less, use
whichever system works better, and don't worry about how many epicycles
"really" exist in "ontological space". Same with sets: AC and GCH make the
theory simpler and easier to use, and they don't lead to inconsistency, so
what's your problem with them? As far as I can tell, you seem to think
that sets have some kind of Real Existence, and that we AC-users, even if
we don't get caught in a contradiction, and telling lies, because our
theorems don't agree with the God-given Reality. Is that it, Bill? This
sounds like some kind of Platonism to me. Platonism is certainly a very
respectable position; lots of people with intellects vastly superior to
mine (Cantor, Goedel, Plato, Maddy) have been Platonists; if you'd just
said your were a Platonist, you'd get no argument from me. The part that
confuses me is how your belief in Platonic Setworld follow from, or even
accords with, your acceptance of Occam's Razor. But this discussion is
getting too way too deep for me; I never studied Metaphysics, so the
explanation will probably be over my head.

> |> It would be desirable to dispense with sets
> |> entirely, since they don't really exist in the physical world.
>
> If I may be permitted, the physical world is almost entirely irrelevant.
> Math is the study of things in the "logical" world, maybe Popper-3, maybe less
> than that. What I called "ontological space". (I really like that term!)
>
> Also, a great many sets *do* exist in this world. But not *all* those in ZF.

What do you have in mind? What kind of sets exist in "this world"? Do you
mean chess sets, sets of dishes, the jet set, and like that?

> |> Occamists only use the constructible sets (and damned few of those, but
>

> Yes, 100%. But we shouldn't even *refer* to the concept constructible sets,
> that's already got a whiff of non-constructibility about it. (That Richard!!)

I don't know much about philosophy or logic or set theory, but I'm an
expert on nonsense, and your statement quoted above is [flamebait deleted]

[Bill Taylor]

Planar <Damien....@see.my.web.page> writes:

|> >but certainly that any set which CANNOT be spoken of, IS NOT. It is silly
|> >even to speak of such, as none can ever be specifically mentioned.
|>

|> But isn't this a bit of circular reasoning ?

Well, in foundations some "circularity" is inevitable. There must be some
bedrock of concept which is unrewritable in simpler terms! Anywhere.

|> How do you define "can
|> be spoken of", independently of the axioms ?

You can't altogether.

|> The goal of the ZF
|> axioms is to define what sets exists, and can be spoken of, isn't it?

Not altogether - it's to ensure that sets that, being speakable-of AND
consistent, actually *do* exist in the theory.

|> As far as I can tell, your line of reasoning could be used to deny the
|> existence of N itself, because it cannot be spoken of, unless you
|> admit the axiom of infinity.

I think I see your problem. The thing is, all the other axioms, (apart from
extensionality which amounts to a *definition* of "set"), are stating the
existence of sets that are already nameable. AxInfinity says N =
{ x | x = 0 or Ey in N: x=s(y)} exists. But note it is *referable to* already.
[0 and s are already simply definable in the theory; the definition is
self-referential, but in a harmless way, so it is quite good from our POV.]

There may be some quibbles about AxFndn, but this is merely technical - of no
concern to math as a whole. The other axioms are definitional. But NOT AC.

-------------------------------------------------------------------------------
Bill Taylor W.Ta...@math.canterbury.ac.nz
-------------------------------------------------------------------------------

There is a fine line between what's negative and what's not.
-------------------------------------------------------------------------------
[PUN ALERT]

[Ilias Kastanas]

Very nice idea -- but the parameterization has problems. For
each point on the x-axis, you have a line; the z-axis, or a line parallel
to the yz plane. The 2nd equation provides lines through any point
(0,1/u,1/u) -- but for each u you have to pick a single v. And worse,
there are no lines through the remaining points of the yz plane -- any
one that doesn't have y = 0 or y = z.

Ilias

[Bill Taylor]

Hopefully this article arrived late at our site, as I've answered most of
these points already, that da...@cogentex.com (Daryl McCullough) writes:

|> that really exist are the definable ones, then the axiom
|> of choice *holds*! You can well-order any collection of
|> sets by well-ordering their definitions.

No. The "observable" well-ordering is OUTSIDE the theory (at best). It cannot
be proved INSIDE. Indeed, the definable sets cannot even be referred to there.

|> that the only sets that exist are the definable ones
|> squares with my intuition of positive real numbers as lengths.

You must be a physicist at heart, Daryl. ;-) Most of us mathies would
say there is little that the physical world can say to influence the
philosophy of math. (Methodology is another matter!)

|> Given a physical object, such as a tree or a
|> book, if you measure its length, I don't see any reason
|> whatsoever (a priori) to think that the result should
|> be a definable real.

OUCH! That's a bit like asking whether the mean distance from the centre
of the earth to the centre of the moon is rational ot irrational. IMHO
these questions are virtually meaningless, for several different reasons -
definitional, verificational, historical, logical, ontological...

|> there is really a limit to the meaningful precision in
|> the length of a physical object. However, I don't see
|> any reason that this must be the case.

This last proviso sounds quite vacuous to me!

|> I definitely
|> don't want to skew mathematics so that our current
|> theory of physics dictates what is mathematically possible.

YES! Agreed. But I would extend it to - any imaginable theory of physics!

|> Third point: I don't understand quite the difference
|> between your qualms about undefinable sets and the
|> intuitionists qualms about unconstructable sets. Could
|> you explain the difference?

An example should be best - again: The set of numbers of nonhalting machines.

This is meaningless to a constructivist; but a perfectly good definition to
a definitionalist, and in fact the definition of a fairly simple set; (and one
with an irreproachable warrant of reality IMHO).

|> What does existence mean for abstractions like sets? You seem to say
|> that things exist if they have names.

That's about the size of it!

|> But, if you just throw in a new
|> choice operator into set theory (that is, a function symbol F with the
|> axiom that all x, (x nonempty implies F(x) is an element of x)) then
|> voila! You have names for all the choice sets.

My very point about the magic function c ! (This *must* be a slow site!)
That's why AC is so silly at heart - it gives a name BY FIAT, rather than
in any natural (much less canonical) way.

Cheers,

-------------------------------------------------------------------------------
Bill Taylor W.Ta...@math.canterbury.ac.nz
-------------------------------------------------------------------------------

G.I.T. for physics?
Well, Godel's proof doesn't seem to carry over.

A closer analogue is that every sufficiently complicated
universe must contain frustrated physicists.

There is some experimental evidence for this.
-------------------------------------------------------------------------------

[Planar]

|> From: Planar <Damien....@see.my.web.page>

|> The goal of the ZF
|> axioms is to define what sets exists, and can be spoken of, isn't it?

>From: mat...@math.canterbury.ac.nz (Bill Taylor)

>Not altogether - it's to ensure that sets that, being speakable-of AND
>consistent, actually *do* exist in the theory.

But "speakable-of" depends on your language and your choice of axioms.
I contend that it is the same as "existing in the theory".

>I think I see your problem. The thing is, all the other axioms, (apart from
>extensionality which amounts to a *definition* of "set"), are stating the

Minor point: I'd say extensionality is the definition of equality,
rather than "set". "Set" belongs to that undefinable bedrock of
concept (obviously). Or if you prefer, the definition of "set" is all
of the axioms together.

>existence of sets that are already nameable. AxInfinity says N =
>{ x | x = 0 or Ey in N: x=s(y)} exists. But note it is *referable to*
>already.

Let S be a set of nonempty sets. C is a choice function iff
C is a set of pairs
and (A(x,y) in C, y in x)
and (A(x,y) in C, A(z,t) in C, x=z => y=t).

So we can speak of choice functions. It seems that you refuse to
admit their existence because I cannot describe a particular one. In
other words, they do not exist because there are too many of them ?

>[0 and s are already simply definable in the theory; the definition is
>self-referential, but in a harmless way, so it is quite good from our POV.]

I don't believe it's quite so harmless. There are a lot of sets that
fit your definition of N above. Which one do you mean ? Is this not
the same problem that I had above for picking one choice function ?

>There may be some quibbles about AxFndn, but this is merely technical - of no
>concern to math as a whole.

About the axiom of foundation, I have pretty much the same question as
you have about AC: what is it used for, exactly ? Which theorems
depend on it ? It seems that Bourbaki don't have an axiom of
foundation, but they say they cover all of mathematics with their
theory. What's going on here ?

> The other axioms are definitional. But NOT AC.

I have a constructivist colleague who doesn't use excluded middle, but
agrees that AC is constructive. (I can elaborate if you wish).

-- Planar

P.S. I'm a formalist. You could say "I want to use AC because that's
what everybody does", or "I want to use not(AC) because it's more
fun", or "I want to use neither so my theorems are compatible with
both theories", all three seem perfectly reasonable to me. But you
cannot say "AC is false", because it was proved to be neither true nor
false (or both, if ZF is inconsistent).

[adam louis stephanides]

mat...@math.canterbury.ac.nz (Bill Taylor) writes:

>Now my "definitionalist" viewpoint is that *any* set that can be spoken of,
>(like the above), probably IS, in some sense (there is room for debate here);

>but certainly that any set which CANNOT be spoken of, IS NOT. It is silly
>even to speak of such, as none can ever be specifically mentioned.

[snip]

>No, I think there are still two confusions here. (i) the confusion between

>constructive and definable (which I have dealt with several times); (ii) the

>confusion between theory and meta-theory (ditto). On the latter, recall

>that the whole universe of actually existing sets MAY be even countable,

>(it may not too, there are subtle points debatable here, also mentioned before);

>but that is COUNTABLE SEEN FROM THE META_THEORY. Internally, in the theory,
>of course there are many uncountable sets. So AC-w is not really relevant.

I've only been following this thread from afar, so to speak, and I admit
I haven't read the Maddy article you talk about, but I really don't
understand this aspect of your position. You seem to take the view that
the universe of sets really exists, from which it would seem to follow
that given a verbal description of a putative set, it
either really corresponds to an actual set, or it doesn't. From this
it would seem to follow that if there really exists a 1-to-1 correspon-
dence between a given set (or the universe) and N, then that set is
really countable, and if there doesn't exist such a 1-to-1 correspondence,
then that set really isn't. I don't see where the concept of a metatheory
fits in this picture.

--Adam