Lebesgue Measure vs. Inaccessible Cardinals

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Jonathan W. Hoyle

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May 7, 1999, 3:00:00 AM5/7/99
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In Ian Stewart's book "From Here to Infinity", the following statement
is made on p. 69:

"...R. M. Solovay showed that the axiom 'there exists an inaccessible
cardinal' implies that every set of reals is Lebesgue measurable...and
subsequently Saharon Shelah proved the converse."

Is this correct? Am I missing something?

Since the Axiom of Choice is all that is required to be able to
construct a non-Lebesgue measurable set, the above quote implies the
statement "the Axiom of Choice is true iff there are no inaccessible
cardinals." Yet in all my readings of inaccessible cardinals, I have
never heard this before.

In fact, in "Infinity and the Mind", Rudy Rucker spends 45 pages in
wonderful detail on the chapter entitled "The Transfinite Cardinals",
and never once mentions this result, despite describing in detail both
the Axiom of Choice and inaccessible cardinals.

Is the term "inaccessible cardinal" being used to describe something
different in Stewart's book than is traditionally meant?

Richard Carr

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May 7, 1999, 3:00:00 AM5/7/99
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On Fri, 7 May 1999, Jonathan W. Hoyle wrote:

:Date: Fri, 07 May 1999 21:52:31 -0400
:From: Jonathan W. Hoyle <jho...@rochester.rr.com>
:Newsgroups: sci.math, sci.logic
:Subject: Lebesgue Measure vs. Inaccessible Cardinals
:
:In Ian Stewart's book "From Here to Infinity", the following statement


:is made on p. 69:
:
:"...R. M. Solovay showed that the axiom 'there exists an inaccessible
:cardinal' implies that every set of reals is Lebesgue measurable...and
:subsequently Saharon Shelah proved the converse."

If this is the exact quote then Stewart is incorrect. Solovay's result is
if ZFC+there exists an inaccessible cardinal is consistent then so is
ZF+a whole load of other things :).
One of these things was that every set of reals is Lebesgue measurable.
Although obviously one does not get AC in the model, one does get
countable choice and indeed a bit more.
Details of this can be found, for example, in A. Kanamori (The Higher
Infinite). It was a very early forcing.

Shelah showed that Con(ZF+every set of reals is Lebesgue measurable)
implies Con(ZF(C?)+there exists an inaccessible cardinal).
The paper you'd need to look at is Sh176.

:
:Is this correct? Am I missing something?

As stated it wasn't. You are alert rather than missing something.

:
:Since the Axiom of Choice is all that is required to be able to


:construct a non-Lebesgue measurable set, the above quote implies the
:statement "the Axiom of Choice is true iff there are no inaccessible
:cardinals." Yet in all my readings of inaccessible cardinals, I have
:never heard this before.

Since it is not possible to prove (in ZF(C)) the consistency of the
existence of inaccessible cardinals but it is possible to prove (in ZF)
the consistency of both ZFC and ZF+not AC, then it is no surprise you
haven't heard it. I think that the consistency of an inaccessible cardinal
also implies the consistency of V=L+there is an inaccessible cardinal or
at least of GCH+there is an inaccessible cardinal. GCH implies AC, V=L
implies a very strong version of AC, so that would totally destroy the
supposed equivalence. (The supposed equivalence also imply that it would
not be possible in ZF to prove ZF+not AC consistent, which certainly is
not the case.)

:
:In fact, in "Infinity and the Mind", Rudy Rucker spends 45 pages in


:wonderful detail on the chapter entitled "The Transfinite Cardinals",
:and never once mentions this result, despite describing in detail both
:the Axiom of Choice and inaccessible cardinals.

This is understandable.
:
:Is the term "inaccessible cardinal" being used to describe something


:different in Stewart's book than is traditionally meant?

I'm not sure, having not read Stewart's book. Generally inaccessible is
either going to mean weakly or strongly inaccessible and usually
inaccessible is reserved just for strongly inaccessible (so that if you
want to talk about weakly inaccessible you add the qualifying "weakly".
The nice thing about kappa being strongly inaccessible is that V_kappa
models ZFC (if V does).
(Under GCH, both notions of inaccessibilty are the same, of course.)

Richard Carr.


Pierre Asselin

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May 8, 1999, 3:00:00 AM5/8/99
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"Jonathan W. Hoyle" <jho...@rochester.rr.com> writes:

>In Ian Stewart's book "From Here to Infinity", the following statement
>is made on p. 69:

>"...R. M. Solovay showed that the axiom 'there exists an inaccessible
>cardinal' implies that every set of reals is Lebesgue measurable...and
>subsequently Saharon Shelah proved the converse."

>Is this correct? Am I missing something?

If the quote is accurate, Stewart goofed. Solovay's result is
that the *consistency* of "there exists an inaccessible cardinal"
implies the *consistency* of "every set of reals is Lebesgue
measurable".

More precisely, let

ZF = Zermelo-Fraenkel
DC = the axiom of dependent choice
IC = there exists an uncountable inaccessible cardinal
LM = every set of reals is Lebesgue measurable

Solovay: Consis(ZF+IC) --> Consis(ZF+DC+LM)
Shelah: Consis(ZF+IC) <-- Consis(ZF+DC+LM)

Reference: Stan Wagon, "The Banach-Tarski Paradox", Cambridge
University Press 1985, ISBN 0-521-45704-1.

--
--Pierre Asselin, Westminster, Colorado
l...@netcom.com

Richard Carr

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May 8, 1999, 3:00:00 AM5/8/99
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On Sat, 8 May 1999, Pierre Asselin wrote:

:Date: Sat, 8 May 1999 03:51:05 GMT
:From: Pierre Asselin <l...@netcom.com>
:Newsgroups: sci.math, sci.logic
:Subject: Re: Lebesgue Measure vs. Inaccessible Cardinals
:


:"Jonathan W. Hoyle" <jho...@rochester.rr.com> writes:
:
:>In Ian Stewart's book "From Here to Infinity", the following statement
:>is made on p. 69:
:
:>"...R. M. Solovay showed that the axiom 'there exists an inaccessible
:>cardinal' implies that every set of reals is Lebesgue measurable...and
:>subsequently Saharon Shelah proved the converse."
:
:>Is this correct? Am I missing something?
:
:If the quote is accurate, Stewart goofed. Solovay's result is
:that the *consistency* of "there exists an inaccessible cardinal"
:implies the *consistency* of "every set of reals is Lebesgue
:measurable".
:
:More precisely, let
:
: ZF = Zermelo-Fraenkel
: DC = the axiom of dependent choice
: IC = there exists an uncountable inaccessible cardinal
: LM = every set of reals is Lebesgue measurable
:
:Solovay: Consis(ZF+IC) --> Consis(ZF+DC+LM)

Indeed it was even more. If B=every set of reals has the Baire property
and PS=every set of reals has the perfect set property then Solovay got
Consis(ZFC+IC)->Consis(ZF+DC+LM+B+PS).
I think he had ZFC+IC on the left, not ZF+IC.

:Shelah: Consis(ZF+IC) <-- Consis(ZF+DC+LM)

:
:


Jonathan W. Hoyle

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May 8, 1999, 3:00:00 AM5/8/99
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Thanks for the feedback, as what you all are saying makes much more
sense. What Stewart apparently meant to say was that inaccessible
cardinals imply *the consistency of* every set of reals being Lebesgue
measurable. This sounds like an editing error in the text, although it
is conceivably possible that the author was just mistaken on this point.

I think I'll email the author about it, so that future printings will
have this fixed. Thanks again to you all.

Mike Oliver

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May 8, 1999, 3:00:00 AM5/8/99
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Richard Carr wrote:
> Indeed it was even more. If B=every set of reals has the Baire property
> and PS=every set of reals has the perfect set property then Solovay got
> Consis(ZFC+IC)->Consis(ZF+DC+LM+B+PS).
> I think he had ZFC+IC on the left, not ZF+IC.

Well, superficially at least there's some tiny question about what
it means to be an inaccessible cardinal in the absence of Choice.
That is, an inaccessible cardinal is one that's regular and strongly
limit, where kappa is strongly limit iff for every lambda < kappa,
2^lambda < kappa.

If we don't have Choice, what does this "<" mean? If A<B means there's
an injection from A to B and none from B to A (the usual meaning without
choice), then if kappa is inaccessible then Choice must hold a fortiori
up to rank kappa. In that case L(V_kappa) is an inner model of
ZFC+"there exists an inaccessible", so we would have Con(ZF+IC) --> Con(ZFC+IC).

Richard Carr

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May 8, 1999, 3:00:00 AM5/8/99
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On Sat, 8 May 1999, Mike Oliver wrote:

:Date: Sat, 08 May 1999 17:43:10 -0700
:From: Mike Oliver <oli...@math.ucla.edu>


:Newsgroups: sci.math, sci.logic
:Subject: Re: Lebesgue Measure vs. Inaccessible Cardinals
:
:

:

:
:

OK.


Mike Oliver

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May 8, 1999, 3:00:00 AM5/8/99
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Richard Carr wrote:

> OK.

Mm, maybe not. I thought about it some more and I'm pretty sure I was wrong
about one part.

Certainly if kappa is inaccessible in the sense given then V_kappa |= AC,
but that doesn't in any obvious way mean that there's a wellordering
of V_kappa itself, particularly one that will show up in L(V_kappa).
In fact I now conjecture the following: If kappa is inaccessible in the
ground model, then there is a forcing notion adding a subset of kappa
such that in the extension, kappa is still inaccessible, but L(V_kappa) |= (not AC).
But I haven't checked this.

However we can still save the conclusion that Con(ZF+IC) --> Con(ZFC+IC).
For, suppose kappa is inaccessible in the sense given. Then surely kappa
is weakly inaccessible: Given lambda<kappa, lambda^+ can be injected into
the set of equivalence classes of wellorderings of lambda, the equivalence
relation being having the same length. So lambda^+ <= 2^2^lambda < kappa.
But now kappa is weakly inaccessible in L, but in L weak inaccessibility
and strong inacessibilty are the same. So L is an inner model of ZFC+IC.

Now there's still one point I haven't touched: I've been implicitly assuming
that an inaccessible cardinal must at least be an *ordinal*. What if we
relax this definition, and say that a set A is inaccessible if

1) it's strongly limit in the sense that for B \subset A there's
an injection from P(B) into A and no injection from A into P(B),

and

2) it's regular in the sense that for any index set I \subset A and any
family {B_i |i \in I} of subsets of A, there's an injection from
(Sigma_{i \in I} B_i) into A, and no injection in the opposite direction
(here Sigma represents the disjoint union).

It's not obvious to me that such an A must be wellorderable (a wellorderable
A would have as its cardinality an inaccessible in the sense that I've been
treating). Does anyone know?

Mike Oliver

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May 8, 1999, 3:00:00 AM5/8/99
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I, Mike Oliver, wrote:
> Now there's still one point I haven't touched: I've been implicitly assuming
> that an inaccessible cardinal must at least be an *ordinal*. What if we
> relax this definition, and say that a set A is inaccessible if
>
> 1) it's strongly limit in the sense that for B \subset A there's
> an injection from P(B) into A and no injection from A into P(B),
>
> and
>
> 2) it's regular in the sense that for any index set I \subset A and any
> family {B_i |i \in I} of subsets of A, there's an injection from
> (Sigma_{i \in I} B_i) into A, and no injection in the opposite direction
> (here Sigma represents the disjoint union).


Oh oh, messed this up too. Of course in clause (1) I want to say |B| < |A| ;
as it stands we could have |B| = |A| which won't do. Similarly in clause
(2) I need |I| < |A| and |B_i| < |A| for each i \in I.

Dr Sinister

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May 9, 1999, 3:00:00 AM5/9/99
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Richard Carr <ca...@math.columbia.edu> wrote in
<Pine.LNX.4.10.990508...@cpw.math.columbia.edu>:

I'm not sure if any of you care to hear this, but so far no one has
addressed this naive issue I posted. Perhaps it will do better in this
thread. Here is what I posted earlier:

--

Dear Webster Kehr,

I have read some of your long paper, and I find it both interesting and
entertaining.

No doubt the following may be a naive question. But, hey, if we don't
ask, we don't learn.

If all the subsets of R(0,1) are countable, then what does that imply for
their measure? Specifically, how do you define a measure such that the
following function f(x) is Lebesgue integrable?

f(x) = 0 if x in R1 = {r1, r2, ...}, rn in R(0,1), R1 countable.
f(x) = 1 if x in complement(R1) wrt R(0,1).

Perhaps Nathan can comment on this as well.


--
No true Scotsman can be an atheist.

ilias kastanas 08-14-90

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May 9, 1999, 3:00:00 AM5/9/99
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In article <3734DA1E...@math.ucla.edu>,
Mike Oliver <oli...@math.ucla.edu> wrote:
@
@Richard Carr wrote:
@> Indeed it was even more. If B=every set of reals has the Baire property
@> and PS=every set of reals has the perfect set property then Solovay got
@> Consis(ZFC+IC)->Consis(ZF+DC+LM+B+PS).
@> I think he had ZFC+IC on the left, not ZF+IC.
@
@Well, superficially at least there's some tiny question about what
@it means to be an inaccessible cardinal in the absence of Choice.
@That is, an inaccessible cardinal is one that's regular and strongly
@limit, where kappa is strongly limit iff for every lambda < kappa,
@2^lambda < kappa.
@
@If we don't have Choice, what does this "<" mean? If A<B means there's
@an injection from A to B and none from B to A (the usual meaning without
@choice), then if kappa is inaccessible then Choice must hold a fortiori
@up to rank kappa. In that case L(V_kappa) is an inner model of
@ZFC+"there exists an inaccessible", so we would have Con(ZF+IC) --> Con(ZFC+IC).

If kappa is weakly inaccessible, then kappa is weakly inaccessible in
L... and hence strongly, too. So L is a model as desired.

Ilias


Richard Carr

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May 9, 1999, 3:00:00 AM5/9/99
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On 9 May 1999, Dr Sinister wrote:

:Date: 9 May 1999 03:50:53 GMT
:From: Dr Sinister <kickass.d...@hotmail.com>


:Newsgroups: sci.math, sci.logic
:Subject: Re: Lebesgue Measure vs. Inaccessible Cardinals
:

:Richard Carr <ca...@math.columbia.edu> wrote in

This is not true.

:<Pine.LNX.4.10.990508...@cpw.math.columbia.edu>:

:
:


Herman Rubin

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May 10, 1999, 3:00:00 AM5/10/99
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In article <3734DA1E...@math.ucla.edu>,
Mike Oliver <oli...@math.ucla.edu> wrote:


>Richard Carr wrote:
>> Indeed it was even more. If B=every set of reals has the Baire property

>> and PS=every set of reals has the perfect set property then Solovay got

>> Consis(ZFC+IC)->Consis(ZF+DC+LM+B+PS).


>> I think he had ZFC+IC on the left, not ZF+IC.

>Well, superficially at least there's some tiny question about what


>it means to be an inaccessible cardinal in the absence of Choice.

>That is, an inaccessible cardinal is one that's regular and strongly

>limit, where kappa is strongly limit iff for every lambda < kappa,

>2^lambda < kappa.

>If we don't have Choice, what does this "<" mean? If A<B means there's

>an injection from A to B and none from B to A (the usual meaning without

>choice), then if kappa is inaccessible then Choice must hold a fortiori

>up to rank kappa. In that case L(V_kappa) is an inner model of

>ZFC+"there exists an inaccessible", so we would have Con(ZF+IC) --> Con(ZFC+IC).

Weakly inaccessible does not involve AC at all; it is the statement
that and ordinal \kappa (larger than \omega) is not the sum of fewer
than \kappa smaller ordinals. Choice is not involved, as the ordinals
are well ordered.

Strong inaccessibility might be a little harder, but I think that
replacing 2^lambda < kappa by H(2^lambda) < kappa, where H is the
Hartogs function, will work adequately.
--
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

Mike Oliver

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May 10, 1999, 3:00:00 AM5/10/99
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Herman Rubin wrote:
> Weakly inaccessible does not involve AC at all;

Yes, but in my usage (which I believe is standard) "inaccessible"
means "strongly inaccessible" unless the opposite is clear from context.

> Strong inaccessibility might be a little harder, but I think that
> replacing 2^lambda < kappa by H(2^lambda) < kappa, where H is the
> Hartogs function, will work adequately.

At least formally there seem to be a number of possibilities; I haven't
worked out all the implications between them. Let's say a set A
is "injectively strongly limit" if for any set B such that there's
an injection from B to A and none from A to B, there's also an injection
from P(B) to A and none from A to P(B).

OTOH A will be "surjectively strongly limit" if for
any B such that there's a surjection from A to B and none from B to A,
there's also a surjection from A to P(B) and none from P(B) to A.

A will be "injectively regular" if given an index set I with |I| < |A|
injectively and sets {B_i | i \in I}, each |B_i| < |A| injectively,
we have that | Sigma_{i \in I} B_i | < |A|, where the cardinalities
are interpreted injectively and Sigma is the disjoint union. I don't
care to define "surjectively regular" at the moment because I'm not
sure whether the cardinality of the index set should be interpreted
injectively or surjectively.

Say A is "injectively inaccessible" if A is injectively regular and
injectively strongly limit. If kappa is an initial ordinal and is
injectively inaccessible, then kappa is inaccessible in the sense
of my recent posts. In that case V_kappa |= AC; this takes a small
argument which I'll post if desired. I still don't know whether
necessarily L(V_kappa) |= AC -- anyone?

If kappa is an initial ordinal then there is no worry about the
definition of regularity, so then we can say kappa is "surjectively
inaccessible" if it's regular and surjectively strongly limit.
I don't know whether this is equivalent to its being injectively
inaccessible, or whether there's an implication in either direction.

I also don't know whether an injectively inaccessible set can
fail to be wellorderable.

The last three paragraphs contain at least three questions; does
anyone know the answers?

--
Disclaimer: I could be wrong -- but I'm not. (Eagles, "Victim of Love")

Finger for PGP public key, or visit http://www.math.ucla.edu/~oliver.
1500 bits, fingerprint AE AE 4F F8 EA EA A6 FB E9 36 5F 9E EA D0 F8 B9

Mike Oliver

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May 10, 1999, 3:00:00 AM5/10/99
to oli...@math.ucla.edu
I, Mike Oliver wrote:
> If kappa is an initial ordinal then there is no worry about the
> definition of regularity, so then we can say kappa is "surjectively
> inaccessible" if it's regular and surjectively strongly limit.
> I don't know whether this is equivalent to its being injectively
> inaccessible, or whether there's an implication in either direction.

OK, got this one. The key is that if alpha is an *ordinal* and A
is a set, then |A| < |alpha| injectively iff |A| < |alpha| surjectively.
That's because either inequality implies that A may be wellordered,
so we can replace A by an initial ordinal.

Therefore *for*initial*ordinals* there's only one notion of inaccessibility.
For non-wellorderable cardinals I still don't know.

(Note that it's *not* true that |alpha| < |A| injectively iff |alpha| < |A|
surjectively. Counterexample: assume AD and let A=R, alpha=omega_1.)

Mike Oliver

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May 10, 1999, 3:00:00 AM5/10/99
to

I, Mike Oliver wrote:

>
> Therefore *for*initial*ordinals* there's only one notion of inaccessibility.
> For non-wellorderable cardinals I still don't know.
>

Spoke too soon once again. Herman's notion (where "kappa strongly limit"
is replaced by "For all lambda<kappa, the Hartogs number of P(lambda) is also
less than kappa") does appear to be different from the notions I was considering.
I was vaguely thinking that it was subsumed in "surjective inaccessibility" but
it's not; to say that the Hartogs number of A is less than kappa
is far weaker than to say there's a surjection from kappa to A and
none in the other direction. (It's equivalent to saying there's a beta<kappa
with no surjection A -->> beta .)

This definition is attractive in a couple of ways: 1) it seems to be
consistent with failures of AC below kappa and 2) I think it relativizes
down to transitive models of ZF.

There's one problem that occurs to me; maybe someone knows a way around
it. The definition as stated says that if lamda<kappa then Theta(P(lambda))<kappa
but it doesn't seem to say anything at all about P(P(lambda)). We could
change the definition to say "If Theta(A)<kappa then Theta(P(A))<kappa", but
now it's no longer obvious that it relativizes down; when you go to the
smaller model there could be more A's such that Theta(A)<kappa .

Randall Dougherty

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May 12, 1999, 3:00:00 AM5/12/99
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In article <3737726B...@math.ucla.edu>,

Mike Oliver <oli...@math.ucla.edu> wrote:
>At least formally there seem to be a number of possibilities; I haven't
>worked out all the implications between them. Let's say a set A
>is "injectively strongly limit" if for any set B such that there's
>an injection from B to A and none from A to B, there's also an injection
>from P(B) to A and none from A to P(B).
>
>A will be "injectively regular" if given an index set I with |I| < |A|
>injectively and sets {B_i | i \in I}, each |B_i| < |A| injectively,
>we have that | Sigma_{i \in I} B_i | < |A|, where the cardinalities
>are interpreted injectively and Sigma is the disjoint union. I don't
>care to define "surjectively regular" at the moment because I'm not
>sure whether the cardinality of the index set should be interpreted
>injectively or surjectively.
>
>Say A is "injectively inaccessible" if A is injectively regular and
>injectively strongly limit. If kappa is an initial ordinal and is
>injectively inaccessible, then kappa is inaccessible in the sense
>of my recent posts. In that case V_kappa |= AC; this takes a small
>argument which I'll post if desired. I still don't know whether
>necessarily L(V_kappa) |= AC -- anyone?

Not necessarily -- one can use symmetric forcing methods (as described
in Chapter 21 of Jech's Set Theory) to construct a counterexample.

Start with the constructible universe L and an inaccessible cardinal
kappa. Consider the Easton forcing construction which adds beta^++
subsets of beta for each regular cardinal beta < kappa; let M be
the resulting generic extension. In M, kappa is still inaccessible.

Let G be the group of automorphisms of the forcing notion which, for
each beta, allows one to permute the beta^++ new subsets of beta
(for all beta simultaneously). For each alpha < kappa, one can consider the
subgroup of G consisting of those permutations which leave fixed the
new subsets of beta for beta < alpha. These subgroups generate a
normal filter; let N be the symmetric submodel of M obtained by using
only hereditarily symmetric names under this filter.

For any alpha < kappa, the part of the M-generic filter giving
just the new subsets of beta for beta < alpha is in N. One can then
show that V_kappa is the same in N as in M. On the other hand,
any subset of kappa in N has a heretitarily symmetric name, and
hence lies in L[S_alpha] for some alpha, where S_alpha is the sequence
of new subsets of beta for beta < alpha, and alpha < kappa.
But any well-ordering of V_kappa can be rearranged to have order
type kappa, and can then be used to get a subset of kappa
encoding _all_ of the new subsets of beta for all beta < kappa;
therefore, N (and hence L(V_kappa)) does not contain a well-ordering
of V_kappa.

>I also don't know whether an injectively inaccessible set can
>fail to be wellorderable.

Symmetric forcing methods work here as well. Start with L and kappa
as above, but this time force to add kappa new subsets of kappa,
with the standard <kappa-closed forcing notion. Let A
be the set of these kappa new sets. Let G be the group
of permutations of kappa (which acts on A in the obvious way),
and for each alpha < kappa, consider the subgroup which fixes the
first alpha elements of A. These again generate a normal filter,
and one gets a symmetric submodel N. Now A is in N, but the
subsets of A in N are precisely those that are bounded
(under the kappa-ordering of A; this ordering is _not_ in N)
and their complements. The complements all have the same
size as A (even in N), and the bounded sets have sizes precisely
all cardinals less than kappa. Hence, A is not well-orderable in N,
but is injectively inaccessible.


Randall Dougherty r...@math.ohio-state.edu
Department of Mathematics, Ohio State University, Columbus, OH 43210 USA
"I have yet to see any problem, however complicated, that when looked at in the
right way didn't become still more complicated." Poul Anderson, "Call Me Joe"

Randall Dougherty

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May 12, 1999, 3:00:00 AM5/12/99
to
In article <7hambh$muv$1...@mathserv.mps.ohio-state.edu>,

Randall Dougherty <r...@math.ohio-state.edu> wrote:
>In article <3737726B...@math.ucla.edu>,
>Mike Oliver <oli...@math.ucla.edu> wrote:
>>I also don't know whether an injectively inaccessible set can
>>fail to be wellorderable.
>
>Symmetric forcing methods work here as well.

However, not the way I just described them. One should be able
to get a model with a set A having the given properties
(all subsets either have the same size as A or have size < kappa;
all sizes less than kappa occur; A is not well-orderable),
but using subsets of kappa in the way described does not work.
Probably the simplest way to do it is to build a permutation model
with the desired properties and apply the Jech-Sochor embedding theorem.
I'll check the details when I get to the office tomorrow.

Nathan the Great

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May 12, 1999, 3:00:00 AM5/12/99
to
Dr. Sinister, are you an evil Cantorian?

> :Dear Webster Kehr,
> :
> :I have read some of your long paper, and I find it both interesting and
> :entertaining.
> :
> :No doubt the following may be a naive question. But, hey, if we don't
> :ask, we don't learn.
> :
> :If all the subsets of R(0,1) are countable, then what does that imply for
> :their measure?

I'm not familiar with the mathematical meaning of "measure". If R[0,1] is cut
into ten equal pieces: [0, .1], [.1, .2], [.2, .3], ... [.9, 1] each piece
will measure 1/10 unit. If R[0,1] is cut into n equal pieces, each piece will
measure 1/n. The sum of these pieces is n*(1/n) = 1. R[0,.1] measures the
same as R(0,.1) because points don't have extension and numbers, as Webster
would say, don't have thickness.

> Specifically, how do you define a measure such that the
> :following function f(x) is Lebesgue integrable?

Can you explain in English?

> :f(x) = 0 if x in R1 = {r1, r2, ...}, rn in R(0,1), R1 countable.
> :f(x) = 1 if x in complement(R1) wrt R(0,1).

How do *you* answer this question?

> :
> :Perhaps Nathan can comment on this as well.
> :

I will be happy to comment, once I figure out what you're talking about.


Nathan

John Savard

unread,
May 12, 1999, 3:00:00 AM5/12/99
to
"Jonathan W. Hoyle" <jho...@rochester.rr.com> wrote, in part:

>Since the Axiom of Choice is all that is required to be able to
>construct a non-Lebesgue measurable set, the above quote implies the
>statement "the Axiom of Choice is true iff there are no inaccessible
>cardinals." Yet in all my readings of inaccessible cardinals, I have
>never heard this before.

From reading the replies to your post, I think your mistake is that
every set of reals could be Lebesgue measurable even if only something
slightly weaker than the Axiom of Choice is true.

Not being an expert on these matters, I made a nontechnical post
("Beyond the Merely Infinite") recently suggesting that perhaps there
are no sets (to which the axiom of the power set applies) whose
cardinality is an uncountable inaccessible cardinal - that while these
numbers could "exist", they would have to be banished from the realm
of set theory to allow the consequences of their existence to be,
somehow, comfortable for me.

Probably just a lucky coincidence, but this seems to indicate my
intuition is working well these days...

John Savard ( teneerf<- )
http://members.xoom.com/quadibloc/index.html

Randall Dougherty

unread,
May 12, 1999, 3:00:00 AM5/12/99
to
In article <7haqp5$ob2$1...@mathserv.mps.ohio-state.edu>,

Randall Dougherty <r...@math.ohio-state.edu> wrote:
>In article <7hambh$muv$1...@mathserv.mps.ohio-state.edu>,
>Randall Dougherty <r...@math.ohio-state.edu> wrote:
>>In article <3737726B...@math.ucla.edu>,
>>Mike Oliver <oli...@math.ucla.edu> wrote:
>>>I also don't know whether an injectively inaccessible set can
>>>fail to be wellorderable.
>>
>>Symmetric forcing methods work here as well.
>
>However, not the way I just described them. One should be able
>to get a model with a set A having the given properties
>(all subsets either have the same size as A or have size < kappa;
>all sizes less than kappa occur; A is not well-orderable),
>but using subsets of kappa in the way described does not work.
>Probably the simplest way to do it is to build a permutation model
>with the desired properties and apply the Jech-Sochor embedding theorem.
>I'll check the details when I get to the office tomorrow.

Yes, this approach via Jech-Sochor does work to give a non-well-orderable
but injectively inaccessible set A. A direct construction
along the lines I gave before also works, but one has to use
sets of new subsets of kappa, not just subsets of kappa, as
elements of the set A.

Mike Oliver

unread,
May 12, 1999, 3:00:00 AM5/12/99
to
Thanks. I see the general outline but I'm missing some details.

Randall Dougherty wrote:
> Start with the constructible universe L and an inaccessible cardinal
> kappa. Consider the Easton forcing construction which adds beta^++
> subsets of beta for each regular cardinal beta < kappa; let M be
> the resulting generic extension. In M, kappa is still inaccessible.

Where do you use the fact that you're increasing the cardinality
of 2^beta? I.e. would everything go through without extra argument
if you simply added *one* subset of beta using <beta-closed forcing,
for each beta<kappa?

> Let G be the group of automorphisms of the forcing notion which, for
> each beta, allows one to permute the beta^++ new subsets of beta
> (for all beta simultaneously).

I could read this as "G = Aut(P), and each element of G leads to permutations
of the new subsets of each beta" or "G is the group of all automorphisms
leading to such permutations". I think you mean the first, right? What
about the *old* subsets of beta? Presumably you could have a name for
a ground-model subset of beta such that *which* subset it names depends
on the generic object. Are such names not hereditarily symmetric?

> For each alpha < kappa, one can consider the
> subgroup of G consisting of those permutations which leave fixed the
> new subsets of beta for beta < alpha. These subgroups generate a
> normal filter; let N be the symmetric submodel of M obtained by using
> only hereditarily symmetric names under this filter.

So such a name is one for which for sufficiently large alpha, if
you fix all new subsets of beta for beta<alpha, then you also fix
the name? I suppose you have to deal with the transitive closure;
does there need to be a name for the transitive closure for which
a *single* alpha works, or can the alpha's be increasing as you
go down the epsilon tree?

> For any alpha < kappa, the part of the M-generic filter giving
> just the new subsets of beta for beta < alpha is in N. One can then
> show that V_kappa is the same in N as in M. On the other hand,
> any subset of kappa in N has a heretitarily symmetric name, and
> hence lies in L[S_alpha] for some alpha, where S_alpha is the sequence
> of new subsets of beta for beta < alpha, and alpha < kappa.

You mean the sequence of sets { {all new subsets of beta | beta<alpha }
or do we also know an ordering for the new subsets of beta for a *particular*
beta?

Bennett Standeven

unread,
May 13, 1999, 3:00:00 AM5/13/99
to

On Wed, 12 May 1999, Nathan the Great wrote:

> Dr. Sinister, are you an evil Cantorian?
>
> > :Dear Webster Kehr,
> > :
> > :I have read some of your long paper, and I find it both interesting and
> > :entertaining.
> > :
> > :No doubt the following may be a naive question. But, hey, if we don't
> > :ask, we don't learn.
> > :
> > :If all the subsets of R(0,1) are countable, then what does that imply for
> > :their measure?
>
> I'm not familiar with the mathematical meaning of "measure". If R[0,1] is cut
> into ten equal pieces: [0, .1], [.1, .2], [.2, .3], ... [.9, 1] each piece
> will measure 1/10 unit. If R[0,1] is cut into n equal pieces, each piece will
> measure 1/n. The sum of these pieces is n*(1/n) = 1. R[0,.1] measures the
> same as R(0,.1) because points don't have extension and numbers, as Webster
> would say, don't have thickness.


Right, so:

1) Points have measure 0.
2) The unit segment has measure 1.

Now to clinch the proof, we need just one more assumption:

3) Measure is countably additive.

IE: If we take the segment (0, 1), the segment (1, 1.5), the segment (1.5,
1.75), etc. and put them all together, we get a set of measure 2.

(The significance of Lebesque measure here is that it has these three
properties)

Then:

Suppose the reals are countable. Then we can express (0,1) as the union of
a countable collection of points. But each point has measure zero, by (1).
Therefore, by (3) the unit interval must have measure 0. But by (2), it
must have measure 1. Therefore, if the reals are countable, then 1=0, and
so the reals are singleton.


Randall Dougherty

unread,
May 13, 1999, 3:00:00 AM5/13/99
to
In article <373A27B4...@math.ucla.edu>,

Mike Oliver <oli...@math.ucla.edu> wrote:
>Thanks. I see the general outline but I'm missing some details.
>
>Randall Dougherty wrote:
>> Start with the constructible universe L and an inaccessible cardinal
>> kappa. Consider the Easton forcing construction which adds beta^++
>> subsets of beta for each regular cardinal beta < kappa; let M be
>> the resulting generic extension. In M, kappa is still inaccessible.
>
>Where do you use the fact that you're increasing the cardinality
>of 2^beta? I.e. would everything go through without extra argument
>if you simply added *one* subset of beta using <beta-closed forcing,
>for each beta<kappa?

Increasing the cardinality of 2^beta isn't important (I just said that
because that version of the forcing is a bit more well-known). I do
need enough new subsets of beta to be able to permute them usefully,
but beta new subsets of beta will suffice (and adding beta new subsets
of beta is equivalent to adding one new subset of beta).

>> Let G be the group of automorphisms of the forcing notion which, for
>> each beta, allows one to permute the beta^++ new subsets of beta
>> (for all beta simultaneously).
>
>I could read this as "G = Aut(P), and each element of G leads to permutations
>of the new subsets of each beta" or "G is the group of all automorphisms
>leading to such permutations". I think you mean the first, right? What
>about the *old* subsets of beta? Presumably you could have a name for
>a ground-model subset of beta such that *which* subset it names depends
>on the generic object. Are such names not hereditarily symmetric?

I think I need to use more notation to clarify this. For each regular
beta < kappa, let P_beta be the forcing notion for adding
beta new subsets of beta: a condition is a partial function from
beta x beta to {0,1} whose domain has size less than beta.
Let P be the Easton product of P_beta for beta < kappa: a condition
is a sequence <p_beta : beta < kappa> where p_beta is in P_beta
and, for each regular gamma <= kappa, the set of beta < gamma
such that p_beta is nonempty has size less than gamma.

For each regular beta, any permutation pi of beta induces an
automorphism pi' of P_beta via the formula
pi'(p)(pi(eta),rho) = p(eta,rho)
for all eta < beta and rho < beta [here "=" means "both are defined with the
same value or both are undefined]. Let G_beta be the set of
all of these automorphisms pi' (so G_beta is a group of automorphisms
of P_beta). This is what I meant by "permuting the new subsets of beta."
(Ground model subsets of beta are not affected. In fact, this isn't
an action on sets at all; its an action on the forcing poset, which
yields an action on _names_. Yes, a name could be a name for
a ground model subset of beta such that which subset it names
depends on the generic object. Such a name could even be hereditarily
symmetric, if it only depends on a small part of the generic object.)

Any sequence <pi'_beta : beta < kappa> with pi'_beta in G_beta
gives an automorphism of the restricted product P; let G be
the group of all such automorphisms.

>> For each alpha < kappa, one can consider the
>> subgroup of G consisting of those permutations which leave fixed the
>> new subsets of beta for beta < alpha. These subgroups generate a
>> normal filter; let N be the symmetric submodel of M obtained by using
>> only hereditarily symmetric names under this filter.
>
>So such a name is one for which for sufficiently large alpha, if
>you fix all new subsets of beta for beta<alpha, then you also fix
>the name? I suppose you have to deal with the transitive closure;
>does there need to be a name for the transitive closure for which
>a *single* alpha works, or can the alpha's be increasing as you
>go down the epsilon tree?

Yes, that's right, and the alpha's can increase as you go down the
epsilon tree. For instance, each of the beta new subsets of beta
has a canonical name which is hereditarily symmetric (just take alpha
to be greater than beta). Then one can form a set whose members are all of
these subsets, for all beta; the canonical name for this is hereditarily
symmetric, because it is itself symmetric with alpha=0 and its
members are symmetric with varying alpha's. (The members of members and so
on are all canonically named ground model sets, hence completely symmetric.)

>> For any alpha < kappa, the part of the M-generic filter giving
>> just the new subsets of beta for beta < alpha is in N. One can then
>> show that V_kappa is the same in N as in M. On the other hand,
>> any subset of kappa in N has a heretitarily symmetric name, and
>> hence lies in L[S_alpha] for some alpha, where S_alpha is the sequence
>> of new subsets of beta for beta < alpha, and alpha < kappa.
>
>You mean the sequence of sets { {all new subsets of beta | beta<alpha }
>or do we also know an ordering for the new subsets of beta for a *particular*
>beta?

Again more notation will make things more precise. For each alpha,
one can naturally decompose P as a product of Q_alpha and Q^alpha,
where Q_alpha is the Easton product of P_beta for beta < alpha
and Q^alpha is the Easton product of P_beta for beta >= alpha.
Then any generic object H on P is a product H_alpha x H^alpha
where H_alpha is generic on Q_alpha and H^alpha is generic
on Q^alpha (even in L[H_alpha]). The canonical name for H_alpha
is hereditarily symmetric. If a subset of kappa in M has a name
which is symmetric at level alpha, then if a forcing condition
p decides whether a certain gamma is in kappa, then the part of p
in Q_alpha also decides whether gamma is in kappa; hence,
the named subset of kappa is in L[H_alpha]. This is what I
meant above when talking about S_alpha (which is supposed to contain
the information needed to reconstruct H_alpha: the list of
beta new subsets of beta for each beta < alpha, in the order specified
by the generic object).

Dr Sinister

unread,
May 14, 1999, 3:00:00 AM5/14/99
to
Nathan the Great <m...@ashland.baysat.net> wrote in
<37398E64...@ashland.baysat.net>:

>Dr. Sinister, are you an evil Cantorian?

No, I'm not an evil Cantorian.

I'm not a mathematician either, so I don't really care what the outcome
of this debate. I don't have any dogma to follow, nor any papers to
defend. And I don't sponge my living off the government like 99% of
mathematicians seem to do.

>> :Dear Webster Kehr,
>> :
>> :I have read some of your long paper, and I find it both interesting
>> :and entertaining.
>> :
>> :No doubt the following may be a naive question. But, hey, if we don't
>> :ask, we don't learn.
>> :
>> :If all the subsets of R(0,1) are countable, then what does that imply
>> :for their measure?
>
>I'm not familiar with the mathematical meaning of "measure". If R[0,1]
>is cut into ten equal pieces: [0, .1], [.1, .2], [.2, .3], ... [.9, 1]
>each piece will measure 1/10 unit. If R[0,1] is cut into n equal
>pieces, each piece will measure 1/n. The sum of these pieces is n*(1/n)
>= 1. R[0,.1] measures the same as R(0,.1) because points don't have
>extension and numbers, as Webster would say, don't have thickness.

Fine. I will use this idea in a moment.

>> Specifically, how do you define a measure such that the
>> :following function f(x) is Lebesgue integrable?
>
>Can you explain in English?

I am not here to write dissertations. Measurable sets are defined in many
books.

>> :f(x) = 0 if x in R1 = {r1, r2, ...}, rn in R(0,1), R1 countable.
>> :f(x) = 1 if x in complement(R1) wrt R(0,1).
>
>How do *you* answer this question?

As far as I know, this integral = 1.

Now, if R1<R is countable, then it has measure zero. Perhaps one can
redefine a measure for R1<R where R is countable and R1 does not
necessarily have measure zero, as you pointed out above.

Fine, if R1=[0,0.1] then you can assign it some 'measure' 1/10 and then
integrate.

Let R1={0<p/q<1: p<<N,q<<N}. This set is countable. The integral of f(x)
over this set is 1 (f as defined above). How do you construct a measure
for this which agrees with the result obtained from Lebesgue integration?

[snip]

>I will be happy to comment, once I figure out what you're talking about.

I am talking about the measurability of subsets of R. This is something
non-cantorians have to define, unless they don't mind dispensing with
integration altogether.

Andrew Boucher

unread,
May 14, 1999, 3:00:00 AM5/14/99
to

----------
In article <8DC625CEEsin...@news.globalserve.net>,
Dr Sinister <drsin...@my-dejanews.com> wrote:

>I am talking about the measurability of subsets of R. This
is something
>non-cantorians have to define, unless they don't mind
dispensing with
>integration altogether.
>

I thought Riemann integration worked just fine.
Measurability (and Lebesgue integration) is needed for the
funny sets.

Mike McCarty

unread,
May 14, 1999, 3:00:00 AM5/14/99
to
In article <7hhchi$mrr$1...@wanadoo.fr>,
Andrew Boucher <Helene....@wanadoo.fr> wrote:
)
)----------
)In article <8DC625CEEsin...@news.globalserve.net>,
)Dr Sinister <drsin...@my-dejanews.com> wrote:
)
)>I am talking about the measurability of subsets of R. This
)is something
)>non-cantorians have to define, unless they don't mind
)dispensing with
)>integration altogether.
)>
)I thought Riemann integration worked just fine.
)Measurability (and Lebesgue integration) is needed for the
)funny sets.

Not really true. The purpose of using Lebesgue integration (or Daniell,
or whatever) is not that it works for "funny sets", because that is not
the common case. The reason is that the "nice" properties one wants it
to have aren't there for the R. Integral.

It's similar to the insistence on actual infinities when dealing with
the integers. We don't *really* need an infinite number of integers.
But if we don't have them, then we can't always form the sum A+B, and
proofs get needlessly complicated with all kinds of provisos and so on
to state the precise conditions under which the conclusion is true.

In the real world, we *don't* need to integrate singular continuous
densities. But if we don't have the tools necessary to handle them,
then our proofs need all kinds of weird add-on assumptions which get in
the way of making progress.

Likewise, when we state some property holds p.p, in the backs of our
minds we know we are really saying it holds *always*, well, except for
some weird cases which don't matter, and we can lump them together that
way. Measure theory lets us do this simplification. Not using it makes
things complicated.

Mike

--
----
char *p="char *p=%c%s%c;main(){printf(p,34,p,34);}";main(){printf(p,34,p,34);}
This message made from 100% recycled bits.
I don't speak for Alcatel <- They make me say that.

Herman Rubin

unread,
May 14, 1999, 3:00:00 AM5/14/99
to
In article <7hhft1$hb0$1...@relay1.dsccc.com>,
Mike McCarty <jmcc...@sun1307.ssd.usa.alcatel.com> wrote:

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

>Not really true. The purpose of using Lebesgue integration (or Daniell,
>or whatever) is not that it works for "funny sets", because that is not
>the common case. The reason is that the "nice" properties one wants it
>to have aren't there for the R. Integral.

There are "reasonable" functions taking on only the values 0 and 1
which are Lebesgue integrable, but not Riemann integrable, and which
cannot be changed on a set of measure 0 to be so integrable. One way
to get such is to express numbers between 0 and 1 to some base b, and
consider sets A_i defined by the digits a_j for j between (i-1)^2 and
i^2 are all 0. Then consider the indicator function of the set of
points belonging to an even number of A_i. Since the measure of A_i
is b^{-(2i-1)}, almost all points belong to only a finite number of
the A_i, so this is defined except for a set of measure 0. But this
set and its complement both have positive measure in any non-trivial
interval.

This is such a function, and when normalized provides an example of a
probability density which is not Riemann integrable.

Andrew Boucher

unread,
May 15, 1999, 3:00:00 AM5/15/99
to

----------
In article <7hhft1$hb0$1...@relay1.dsccc.com>,
jmcc...@sun1307.ssd.usa.alcatel.com (Mike McCarty) wrote:


>In article <7hhchi$mrr$1...@wanadoo.fr>,
>Andrew Boucher <Helene....@wanadoo.fr> wrote:
>)
>)----------
>)In article
<8DC625CEEsin...@news.globalserve.net>,
>)Dr Sinister <drsin...@my-dejanews.com> wrote:
>)
>)>I am talking about the measurability of subsets of R.
This
>)is something
>)>non-cantorians have to define, unless they don't mind
>)dispensing with
>)>integration altogether.
>)>
>)I thought Riemann integration worked just fine.
>)Measurability (and Lebesgue integration) is needed for the
>)funny sets.
>

>Not really true. The purpose of using Lebesgue integration
(or Daniell,
>or whatever) is not that it works for "funny sets", because
that is not
>the common case. The reason is that the "nice" properties
one wants it
>to have aren't there for the R. Integral.
>

>It's similar to the insistence on actual infinities when
dealing with
>the integers. We don't *really* need an infinite number of
integers.
>But if we don't have them, then we can't always form the
sum A+B, and
>proofs get needlessly complicated with all kinds of
provisos and so on
>to state the precise conditions under which the conclusion
is true.

No agree. First, we do really need an infinite number of
natural numbers, because otherwise I would have to assume
the world is necessarily finite, and I see no reason to
suppose this is the case. Finiteness even strikes me as
incoherent. If n is the largest natural number, isn't {me} U
{1..n} of size n+1? (You *do* have to suppose that I am not
a natural number!) Secondly, even if we
*didn't need* an infinite number of natural numbers, I would
still say my intuition believes in them, so I would assume
it in any case. (Which do *you* prefer? And by that, not
which do you prefer to use, but which do you think of as
actually existing? An unbounded or a bounded set of natural
numbers?)

Which is not to say that your point doesn't show insight,
only that I wouldn't use little itty-bitty trustworthy N to
back it up. I think this type of argument has more force if
you would use the real numbers R. Do we really "need" them?
We do, but only because they make things much, much
simpler.

But just because we make the leap once, doesn't mean we
should always make the leap. Each time should be
scrutinized. And ultimately I think even the analogy with R
breaks down. R is used by the physicist and the engineer.
The guys that build bridges use R. I don't think they're
using non-Riemann integration.

>
>In the real world, we *don't* need to integrate singular
continuous
>densities.

We agree.

>But if we don't have the tools necessary to handle them,
>then our proofs need all kinds of weird add-on assumptions
which get in
>the way of making progress.

I guess we agree that, if you have all kinds of weird
add-ons, you can take that as an indication that you're on
the wrong track. We don't agree whether the conclusion
should be: accept measurability a la Lebesgue; or take
another look at what you're proving and why. Maybe the
add-ons aren't as weird as all that, or maybe they reveal
something about reality. Maybe we don't even need the
proposition to begin with. I don't know how to begin
analysis without the real numbers; but I can get a good way
without Lebesgue integration. After all, we didn't have it
until a hundred years ago...

Mike Oliver

unread,
May 15, 1999, 3:00:00 AM5/15/99
to

Andrew Boucher wrote:
> No agree. First, we do really need an infinite number of
> natural numbers, because otherwise I would have to assume
> the world is necessarily finite, and I see no reason to
> suppose this is the case.

Do you have any reason to suppose it's *not* the case? Besides
the fact that it makes things easier?

But in any case, you don't need to *have* infinitely many naturals
in order to avoid *assuming* a necessarily finite world; you can
simply remain agnostic on whether the world is necessarily finite.

> Finiteness even strikes me as
> incoherent. If n is the largest natural number, isn't {me} U
> {1..n} of size n+1?

How do you know you can form this set?

> Secondly, even if we
> *didn't need* an infinite number of natural numbers, I would
> still say my intuition believes in them, so I would assume
> it in any case.

My intuition has no problems with V.

> I don't know how to begin
> analysis without the real numbers; but I can get a good way
> without Lebesgue integration. After all, we didn't have it
> until a hundred years ago...

Which would be a stronger point if not for the fact that most of
the best mathematics is less than a hundred years old.

Andrew Boucher

unread,
May 16, 1999, 3:00:00 AM5/16/99
to

----------
In article <373E12CE...@math.ucla.edu>, Mike Oliver
<oli...@math.ucla.edu> wrote:


>
>
>Andrew Boucher wrote:
>> No agree. First, we do really need an infinite number of
>> natural numbers, because otherwise I would have to assume
>> the world is necessarily finite, and I see no reason to
>> suppose this is the case.
>
>Do you have any reason to suppose it's *not* the case?
Besides
>the fact that it makes things easier?
>
>But in any case, you don't need to *have* infinitely many
naturals
>in order to avoid *assuming* a necessarily finite world;
you can
>simply remain agnostic on whether the world is necessarily
finite.

No agree. If it's possible the world is infinite, I want an
infinite set of naturals. Just in case.

>
>> Finiteness even strikes me as
>> incoherent. If n is the largest natural number, isn't
{me} U
>> {1..n} of size n+1?
>
>How do you know you can form this set?

Well gee whiz. To be equally flippant, what about the union
axiom? (I look deep inside myself, and I know {me} exists.
The set of naturals exist, and since this set = {1..n}, then
{1..n} exists. Ok so I forgot 0...)

>
>> Secondly, even if we
>> *didn't need* an infinite number of natural numbers, I
would
>> still say my intuition believes in them, so I would
assume
>> it in any case.
>
>My intuition has no problems with V.

Last time I checked, that was another thread...

>
>> I don't know how to begin
>> analysis without the real numbers; but I can get a good
way
>> without Lebesgue integration. After all, we didn't have
it
>> until a hundred years ago...
>
>Which would be a stronger point if not for the fact that
most of
>the best mathematics is less than a hundred years old.

Which would be a stronger point if beauty wasn't in the eye
of the beholder.

Mike Oliver

unread,
May 16, 1999, 3:00:00 AM5/16/99
to
In article <7hlq41$p4$1...@wanadoo.fr>,

"Andrew Boucher" <Helene....@wanadoo.fr> wrote:
>
> ----------
> In article <373E12CE...@math.ucla.edu>, Mike Oliver
> <oli...@math.ucla.edu> wrote:
>> But in any case, you don't need to *have* infinitely many naturals
>> in order to avoid *assuming* a necessarily finite world; you can
>> simply remain agnostic on whether the world is necessarily
>> finite.
>
> No agree. If it's possible the world is infinite, I want an
> infinite set of naturals. Just in case.

But you haven't demonstrated that it's possible that the world
is infinite. What if the assumption that the world is infinite
leads to a contradiction? In that case it's *impossible* that
the world is infinite. And your failure to find this contradiction
doesn't demonstrate that it's not there.

>>> Finiteness even strikes me as
>>> incoherent. If n is the largest natural number, isn't {me} U
>>> {1..n} of size n+1?

>>How do you know you can form this set?

> Well gee whiz. To be equally flippant, what about the union
> axiom? (I look deep inside myself, and I know {me} exists.
> The set of naturals exist, and since this set = {1..n}, then
> {1..n} exists. Ok so I forgot 0...)

I wasn't being flippant. What *about* the union axiom? I don't
claim there's any certainty there. You're the one (apparently)
making claims of certainty. I claim only knowledge in a pragmatic
sense, on which level the mental operations which justify arbitrarily-
large integers do not appear all that different from the ones
which justify, say, supercompact cardinals.


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May 16, 1999, 3:00:00 AM5/16/99