"How bad can 2^(Aleph_0) be?" history question
Dave L. Renfro
with ZFC in 1963. [Kanamori, "The Higher Infinite", p. 114, says
that Cohen circulated notes for the proof in April 1963.]
I am interested in who, when, and where (seminar presentation,
published paper, etc.) proved the stronger result that it is
consistent with ZFC that 2^(Aleph_0) can be any specified cardinal
whose cofinality is greater than Aleph_0. I know that stronger
results were proved by Richard Soloay and William Easton, but I'm
only interested in the possibilities for 2^(Aleph_0).
I think maybe this was proved independently by Cohen and Solovay.
In abstract 63T-395 on page 595 of the Oct. 1963 Notices of the
American Mathematical Society (dated Sept. 3, 1963), Solovay writes
"(The first was discovered independently by Cohen.)" in reference
to the fact that 2^(Aleph_0) can be any cardinal with cofinality
greater than Aleph_0. That Cohen proved this appears to be
confirmed by Lemma 16 on page 49 of his paper "Independence results
in set theory", which appears in John W. Addison, Leon Henkin, and
Alfred Tarski, "The Theory of Models", Proceedings of the 1963
International Symposium at Berkeley, North-Holland, 1965.
However, Cohen mentions a result of Solovay's [Not the result that
I'm interested in, but rather that 2^(Aleph_n) = Aleph_g(n) can
be satisfied simultaneously for all positive integers n in some
model for ZFC, where g is any specified strictly increasing function
from the positive integers into the positive integers.] as well
as a result *later* announced by Easton. Because of this, I would
guess that Cohen wrote this paper well after his Berkeley talk,
and so not all of the results mentioned in it may have been
presented at the Berkeley conference (in particular, Cohen's
Lemma 16 that I mentioned above).
Can someone tell me what dates this conference took place?
I forgot to write this down when I was photocopying this stuff
yesterday and the library I was at is a 160 mile round-trip drive.
Also, does anyone know whether or not Cohen had actually proved the
result I'm interested in--that 2^(Aleph_0) can be any cardinal
with uncountable cofinality--when this conference took place?
If not, when did Cohen obtain it?
Dave L. Renfro
Nico Benschop
"Dave L. Renfro" wrote:
>
> Paul Cohen proved that 2^(Aleph_0) =/= Aleph_1 is consistent
> with ZFC in 1963. [Kanamori, "The Higher Infinite", p. 114, says
> that Cohen circulated notes for the proof in April 1963.]
>
> I am interested in who, when, and where (seminar presentation,
> published paper, etc.) proved the stronger result that it is
> consistent with ZFC that 2^(Aleph_0) can be any specified cardinal
> whose cofinality is greater than Aleph_0. I know that stronger
> results were proved by Richard Soloay and William Easton, but I'm
> only interested in the possibilities for 2^(Aleph_0).
I can't help you with that reference, sorry.
But regarding 2^N, just for argument's sake have a look at:
http://www.com2com.ru/alexzen/papers/Cantor/Fatal_Mistake_of_Cantor.html
by prof. Alex Zenkin (Russian Academy of Sciences) who has some
critique on Cantor's method (if not on its interpretation;-)
-- NB -- http://home.iae.nl/users/benschop/cantor.htm
http://home.iae.nl/users/benschop/seqdim.htm
Robin Chapman
Here, "has some critique" means "spouts some bollocks".
------------------------------------------------------------
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
"His mind has been corrupted by colours, sounds and shapes."
The League of Gentlemen
Wilbert Dijkhof
> >===== Original Message From Nico Benschop <n.ben...@chello.nl> =====
> >"Dave L. Renfro" wrote:
> >>
> >> Paul Cohen proved that 2^(Aleph_0) =/= Aleph_1 is consistent
> >> with ZFC in 1963. [Kanamori, "The Higher Infinite", p. 114, says
> >> that Cohen circulated notes for the proof in April 1963.]
> >>
> >> I am interested in who, when, and where (seminar presentation,
> >> published paper, etc.) proved the stronger result that it is
> >> consistent with ZFC that 2^(Aleph_0) can be any specified cardinal
> >> whose cofinality is greater than Aleph_0. I know that stronger
> >> results were proved by Richard Soloay and William Easton, but I'm
> >> only interested in the possibilities for 2^(Aleph_0).
> >> [...] -- Dave L. Renfro
> >
> >I can't help you with that reference, sorry.
> >But regarding 2^N, just for argument's sake have a look at:
> >
> >c
> >
> >by prof. Alex Zenkin (Russian Academy of Sciences) who has some
> >critique on Cantor's method (if not on its interpretation;-)
>
> Here, "has some critique" means "spouts some bollocks".
Indeed, for example:
"It is obvious that if the enumeration (1) is finite, then its diagonal (*),
and the
corresponding "anti-diagonal" sequence (**) will also be finite, i.e., in
such the
case it does not define any real number. In other words, the inapplicability
of
Cantor's diagonal method to finite enumerations is so obvious that nobody
ever
examined that issue. However, that "obviousness" is only an ingrained
psychological
stereotype of meta-mathematical thinking."
This enumeration is the enumeration of all real numbers. Of course can't the
enumeration be finite. But even if the corresponding diagonal number is
finite it will define a real number.
Wilbert
David C. Ullrich
His "critique" is nonsense. First, he does not even state the
traditional proof properly - he uses reals "base 2", and he doesn't
ever show that the diagonal sequence is not just another
representation for a number already on the list. It's well-known
to everyone but him that base 2 does not work as well as other
bases (the point being that if we're giving the proof with real
numbers represented as "base b decimals" then when we construct
the diagonal sequence we want to make certain that we do not use
0 or b-1 as digits - this means we avoid numbers that have more
than one representation in base b, so the fact that we get a new
sequence of digits really does mean that we got a new real number.
Now if b = 2 this means we cannot use 0 or 1 as digits for the
diagonal number...)
That's details. More important, he simply does not seem to understand
the _logical_ _structure_ of the argument. He says, roughly, "ok, we
have this sequence of reals x_1, x_2,... and we construct a real y
not on the list. So what? Now consider the sequence y, x_1, x_2,..."
Honest. It's exactly the same pathetic misunderstanding behind
half of the proofs that Cantor was Wrong that we see here on
sci.math.
Theorem No countable "list" of reals contains them all.
Proof: Suppose that x_1, x_2, ... is a counatble list of
reals. Do this: [insert construction of y]. Now y is a real,
and y is not one of the x_j, QED.
Arguments about the sequence y, x_1, x_2,... are just stupid;
they indicate that the arguer doesn't understand simple logic.
If you want to show that no X has property P and also property
Q, a valid way to do it is this:
Assume that X has property P.
Show that X does not have property Q.
If you've done that then you _have_ shown that
_no_ X has properties P and Q. It suffices in the
second line to show that _X_ does not have property
Q - the idiotic refutations we sometimes see sound as
though this were required:
Assume that X has property P.
Show that no X' has property Q.
Thinking that that's what's required to show that
no X has both property P and Q is just cloth-headed
nonsense - being a member of the Russian Academy and
including quotes from Aristotle in Latin(!) doesn't
change that fact.
Did you actually think his "critique" must make some
sense because of his credentials? That would be surprising,
coming from a Believe Nobody guy like you. Or did you
actually _read_ it and decide that it made sense or what?
>-- NB -- http://home.iae.nl/users/benschop/cantor.htm
> http://home.iae.nl/users/benschop/seqdim.htm
David C. Ullrich
*********************
"Sometimes you can have access violations all the
time and the program still works." (Michael Caracena,
comp.lang.pascal.delphi.misc 5/1/01)
David C. Ullrich
This is just a silly irrelevancy. The actual nonsense comes
later in the "paper".
> But even if the corresponding diagonal number is
>finite it will define a real number.
>
>Wilbert
>
>
>
David C. Ullrich
Nico Benschop
>
> On Tue, 29 May 2001 10:56:33 +0200, Nico Benschop
> <n.ben...@chello.nl> wrote:
>
> >
> >"Dave L. Renfro" wrote:
> >>
> >> Paul Cohen proved that 2^(Aleph_0) =/= Aleph_1 is consistent
> >> with ZFC in 1963. [Kanamori, "The Higher Infinite", p. 114, says
> >> that Cohen circulated notes for the proof in April 1963.]
> >>
> >> I am interested in who, when, and where (seminar presentation,
> >> published paper, etc.) proved the stronger result that it is
> >> consistent with ZFC that 2^(Aleph_0) can be any specified cardinal
> >> whose cofinality is greater than Aleph_0. I know that stronger
> >> results were proved by Richard Soloay and William Easton, but I'm
> >> only interested in the possibilities for 2^(Aleph_0).
> >> [...] -- Dave L. Renfro
> >
> >I can't help you with that reference, sorry.
> >But regarding 2^N, just for argument's sake have a look at:
> >
> >http://www.com2com.ru/alexzen/papers/Cantor/Fatal_Mistake_of_Cantor.html
> >
> >by prof. Alex Zenkin (Russian Academy of Sciences) who has some
> >critique on Cantor's method (if not on its interpretation;-)
>
> His "critique" is nonsense. First, he does not even state the
> traditional proof properly - he uses reals "base 2", and he doesn't
> ever show that the diagonal sequence is not just another
> representation for a number already on the list. It's well-known
> to everyone but him that base 2 does not work as well as other bases
>
> That's details. More important, he simply does not seem to understand
> the _logical_ _structure_ of the argument. He says, roughly, "ok, we
> have this sequence of reals x_1, x_2,... and we construct a real y
> not on the list. So what? Now consider the sequence y, x_1, x_2,..."
> Honest. It's exactly the same pathetic misunderstanding behind
> half of the proofs that Cantor was Wrong that we see here on
> sci.math. Theorem No countable "list" of reals contains them all.
> Proof: Suppose that x_1, x_2, ... is a counatble list of
> reals. Do this: [insert construction of y]. Now y is a real,
> and y is not one of the x_j, QED.
> Arguments about the sequence y, x_1, x_2,... are just stupid;
> [...]
> no X has both property P and Q is just cloth-headed
> nonsense - being a member of the Russian Academy and
> including quotes from Aristotle in Latin(!) doesn't
>
> Did you actually think his "critique" must make some
> sense because of his credentials? That would be surprising,
> coming from a Believe Nobody guy like you. Or did you
> actually _read_ it and decide that it made sense or what?
>
> >-- NB -- http://home.iae.nl/users/benschop/cantor.htm
> > http://home.iae.nl/users/benschop/seqdim.htm
Don't get excited, David;-)
I did not say I agree with his analysis, but just "for the sake
of argument" to have a look... Now his objection against ONE counter
example (a real not on any initial countable list of length w)
reminded me of my own initial reaction (way back;-) :: so what...?
Instead of w reals now we have w+1 reals, which is _still_ countable.
Clearly ONE new real outside the list suffices logically (copying our
intuition from finite lists;-) to refute the statement that the initial
list was countable. No problem with me.
On the other hand, many people - at first hearing the argument -
must have the nagging feeling of "vanishing evidence": w+1 = w (!)
For a more constructive proof, or rather a more convincing proof,
for those who are not used to strict logic (;-) it might be useful
to point out (as I do in my homepage */cantor.htm ) that not just
one but MANY reals can be shown to be outside the (any!) list...
And that in fact the diagonal argument can be used as a constructive
GENERATIVE device to produce some w! (omega factorial) reals outside
the (any) list: namely by considering all row permutations and the
corresponding diagonals. This is clearly overkill, I admit, but OTOH
it shows that indeed (?) uncountably many reals - in the order of w! -
are outside any list you may have. Note that a countable number of
countable reals outside the list, by repeated extension of it, as
A.Zenkin suggests ('recursive extension') does NOT help, since that
would only yield w x w = w^2 = w (again countable;-(
My point: can it be shown that indeed w! (by permuting a countable
length initial listing) produces "UNcountably" many reals on [0,1) ?
In other words, interpreting UNcountable not so much as "not countable"
(a negative property), but as "ESSENTIALLY more than countable" ...
You get my drift? This is my "sequential dimension" approach via
infinite group Z! (all permutations of the integers Z), which contains
2^Z (all subsets of the integers) ... yet Z! has ONLY 2 generators -
which is ESSENTIALLY another ballpark alltogether: "NOT countable"
requires more than ONE single generator (re: Peano's N = {+1*} iterated
successor function), so at least TWO generators: group Z! containing
Boolean Lattice 2^Z (the set of all fixed-points[-sets] of the
permutations of Z). Think in closures, with a generating operation,
rather than just infinite sets (which really have no 'structure';-)
Roughly, I guess this has to do with refusing to be convinced by a
single (although constructed) counter example in an infinite context:
MUCH more can be said in Cantor's diagonal case! Don't you agree?
(would this have anything to do with my countryman Brouwer?-)
---- One is Always Halfway Anyway --- AHA
xxxxxxxxxxxxxxxxxxx1.1xxxxxxxxxxxxxxxxxxx
David C. Ullrich
wrote:
The idea that that last sentence has any relevance to the proof is
ridiculous.
>Clearly ONE new real outside the list suffices logically (copying our
>intuition from finite lists;-) to refute the statement that the initial
>list was countable. No problem with me.
>
>On the other hand, many people - at first hearing the argument -
>must have the nagging feeling of "vanishing evidence": w+1 = w (!)
>
>For a more constructive proof, or rather a more convincing proof,
>for those who are not used to strict logic (;-) it might be useful
>to point out (as I do in my homepage */cantor.htm ) that not just
>one but MANY reals can be shown to be outside the (any!) list...
Anyone who finds this "more convincing" is very confused. For
that matter the idea that one valid proof can be more convincing
than another valid proof is sort of nonsense as well. One proof
can be more illuminating than another, or easier to follow. But
the original proof in _this_ case is so utterly simple that if
someone finds a more complicated argument "more convincing" that
person has simply not understood the original.
>And that in fact the diagonal argument can be used as a constructive
>GENERATIVE device to produce some w! (omega factorial) reals outside
>the (any) list: namely by considering all row permutations and the
>corresponding diagonals. This is clearly overkill, I admit, but OTOH
>it shows that indeed (?) uncountably many reals - in the order of w! -
>are outside any list you may have. Note that a countable number of
>countable reals outside the list, by repeated extension of it, as
>A.Zenkin suggests ('recursive extension') does NOT help, since that
>would only yield w x w = w^2 = w (again countable;-(
>
>My point: can it be shown that indeed w! (by permuting a countable
>length initial listing) produces "UNcountably" many reals on [0,1) ?
>In other words, interpreting UNcountable not so much as "not countable"
>(a negative property), but as "ESSENTIALLY more than countable" ...
>You get my drift? This is my "sequential dimension" approach via
>infinite group Z! (all permutations of the integers Z), which contains
>2^Z (all subsets of the integers) ... yet Z! has ONLY 2 generators -
>which is ESSENTIALLY another ballpark alltogether: "NOT countable"
>requires more than ONE single generator (re: Peano's N = {+1*} iterated
>successor function), so at least TWO generators: group Z! containing
>Boolean Lattice 2^Z (the set of all fixed-points[-sets] of the
>permutations of Z). Think in closures, with a generating operation,
>rather than just infinite sets (which really have no 'structure';-)
>
>Roughly, I guess this has to do with refusing to be convinced by a
>single (although constructed) counter example in an infinite context:
Yes, it does have to do with refusing to accept one counterexample.
To put it differently, it has to do with stupidity. Cloth-headedness.
What's that word Chapman uses? Bollocks, that's it. (I'm not sure
what a bollock is, but I'm pretty sure I understand what it means
to say a proof is boolocks.)
>MUCH more can be said in Cantor's diagonal case! Don't you agree?
Of course much more can be said. Any time anyone says _anything_
there's _always_ much more that can be said. The idea that much
more _needs_ to be said is nonsense.
>(would this have anything to do with my countryman Brouwer?-)
Nothing as far as I can see. Did you have anything specific in
mind?
>-- NB -- http://home.iae.nl/users/benschop/cantor.htm
> http://home.iae.nl/users/benschop/seqdim.htm
>
>
>
>---- One is Always Halfway Anyway --- AHA
>xxxxxxxxxxxxxxxxxxx1.1xxxxxxxxxxxxxxxxxxx
David C. Ullrich
ma...@primrose.csv.warwick.ac.uk
ull...@math.okstate.edu writes:
<Much omitted>
>To put it differently, it has to do with stupidity. Cloth-headedness.
>What's that word Chapman uses? Bollocks, that's it. (I'm not sure
>what a bollock is, but I'm pretty sure I understand what it means
>to say a proof is boolocks.)
"Bollock" is a slightly vulgar word for testicle - very common (in both
senses of the word) in UK English. Possibly it is derived from the word
"ball", with which it can be used more or less interchangeably.
I am so confident that you have correctly understood what it means to say that
a proof, or anything else for that matter, is bollocks, that I will not take
the trouble to explain.
Derek Holt.
Phil Carmody
>
> In article <3b14f7a0...@nntp.sprynet.com>,
> ull...@math.okstate.edu writes:
>
> <Much omitted>
>
> >To put it differently, it has to do with stupidity. Cloth-headedness.
> >What's that word Chapman uses? Bollocks, that's it. (I'm not sure
> >what a bollock is, but I'm pretty sure I understand what it means
> >to say a proof is boolocks.)
>
> "Bollock" is a slightly vulgar word for testicle - very common (in both
> senses of the word) in UK English. Possibly it is derived from the word
> "ball", with which it can be used more or less interchangeably.
The controvercially named "Never mind the bollocks, here's the sex
pisotls" album weaseled its way out of a court case due to the discovery
of an arcane use of the word "bollock" to mean a church order, some kind
of priest, effectively.
The root of "bollock", "ball", "bolus", and "bullet" in English are all
the same - the Latin for ball.
Dave L. Renfro
[sci.math Tue, 29 May 2001 18:19:08 GMT]
<http://forum.swarthmore.edu/epigone/sci.math/stantaxfan>
wrote (in part)
> For a more constructive proof, or rather a more convincing proof,
> for those who are not used to strict logic (;-) it might be useful
You want to be careful using a somewhat standard math term (even
if many variations of its meanings are used)--constructive--for its
"common language" use, at least in sci.math. This is like saying
"I'm going to work really hard and use all my power to force
this rock to move" in the middle of a discussion on the foundations
of mechanics!
> And that in fact the diagonal argument can be used as a
> constructive GENERATIVE device to produce some w! (omega factorial)
> reals outside the (any) list: namely by considering all row
> permutations and the corresponding diagonals. This is clearly
> overkill, I admit, but OTOH it shows that indeed (?) uncountably
> many reals - in the order of w! - are outside any list you may
> have. Note that a countable number of countable reals outside the
O-K, I'll bite--what's the definition of (w!)? I know what the
ORDINAL number w! is (it's ordinal number w^2), but the only
reasonable definition that I know of for the CARDINAL w! is the
infinite cardinal product 1*2*3*4*..., which is the cardinal w.
Maybe you're thinking of blue jays?
Oh, and if you define the cardinal w! by using a specific countable
set for w, make sure that your definition is well-defined. That is,
your definition outputs the same cardinal number no matter what
countable set is used.
Lastly, shouldn't ANY countable list of real numbers omit
uncountably many (in fact, more than this -- co-countably many)
real numbers? Isn't this trivial?
Dave L. Renfro
Dave L. Renfro
[sci.math 30 May 01 10:33:21 -0400 (EDT)]
<http://forum.swarthmore.edu/epigone/sci.math/stantaxfan>
wrote (in part)
> O-K, I'll bite--what's the definition of (w!)? I know what the
> ORDINAL number w! is (it's ordinal number w^2),
Actually, this might not be all that standard. Sierpinski ("Cardinal
and Ordinal Numbers", 1965 edition, p. 318) defines
Gamma(a) = Product of all ordinals less than a
and
a! = Gamma(a+1).
Hence, Gamma(w) = w and w! = w^2. Nonetheless, both of these are
countable ordinals (extremely tiny countable ones, at that), and
I can't think of any reasonable alternative for defining the
ordinal w!.
By the way, sci.math readers who eagerly follow all of my posts with
breathless anticipation will be thrilled to know that I did wind
up getting a copy of Sierpinski's book, which I mentioned I might at
<http://forum.swarthmore.edu/epigone/sci.math/zuyolquim/ejz42z...@forum.mathforum.com>.
Dave L. Renfro
David C. Ullrich
wrote:
>Nico Benschop <n.ben...@chello.nl>
>[sci.math Tue, 29 May 2001 18:19:08 GMT]
><http://forum.swarthmore.edu/epigone/sci.math/stantaxfan>
>
>wrote (in part)
>
>> For a more constructive proof, or rather a more convincing proof,
>> for those who are not used to strict logic (;-) it might be useful
>
>You want to be careful using a somewhat standard math term (even
>if many variations of its meanings are used)--constructive--for its
>"common language" use, at least in sci.math. This is like saying
>"I'm going to work really hard and use all my power to force
>this rock to move" in the middle of a discussion on the foundations
>of mechanics!
>
>> And that in fact the diagonal argument can be used as a
>> constructive GENERATIVE device to produce some w! (omega factorial)
>> reals outside the (any) list: namely by considering all row
>> permutations and the corresponding diagonals. This is clearly
>> overkill, I admit, but OTOH it shows that indeed (?) uncountably
>> many reals - in the order of w! - are outside any list you may
>> have. Note that a countable number of countable reals outside the
>
>O-K, I'll bite--what's the definition of (w!)? I know what the
>ORDINAL number w! is (it's ordinal number w^2), but the only
>reasonable definition that I know of for the CARDINAL w! is the
>infinite cardinal product 1*2*3*4*..., which is the cardinal w.
>Maybe you're thinking of blue jays?
Not that... never mind the disclaimers. Surely a reasonable
definition for the factorial of a cardinal would be the number
of permutations of a set with that cardinality.
>Oh, and if you define the cardinal w! by using a specific countable
>set for w, make sure that your definition is well-defined. That is,
>your definition outputs the same cardinal number no matter what
>countable set is used.
>
>Lastly, shouldn't ANY countable list of real numbers omit
>uncountably many (in fact, more than this -- co-countably many)
>real numbers? Isn't this trivial?
No, this is very very deep.
>Dave L. Renfro
Nico Benschop
>
> Dave L. Renfro <ren...@central.edu>
> [sci.math 30 May 01 10:33:21 -0400 (EDT)]
> <http://forum.swarthmore.edu/epigone/sci.math/stantaxfan>
>
> wrote (in part)
>
> > O-K, I'll bite--what's the definition of (w!)? I know what the
> > ORDINAL number w! is (it's ordinal number w^2),
>
> Actually, this might not be all that standard. Sierpinski ("Cardinal
> and Ordinal Numbers", 1965 edition, p. 318) defines
>
> Gamma(a) = Product of all ordinals less than a
>
> and
>
> a! = Gamma(a+1).
>
> Hence, Gamma(w) = w and w! = w^2. Nonetheless, both of these are
> countable ordinals (extremely tiny countable ones, at that), and
> I can't think of any reasonable alternative for defining the
All I suggest is the way to generate such infinity (is that OK?-):
how about w! as the cardinality of Z!, the set of all permutations of Z
(the integers). Similar to w as the cardinality of all positive
integers,
and 2^w that of the powerset 2^N of all subsets of N (or of 2^Z, re: Z)
Does'nt it make sense then to compare: w < 2^w < w! ...[1]
corresponding to closures: Z , 2^Z , Z!
(in [1]: possibly the rightmost '<' to be '=')
with Z as cyclic group with _one_ generator,
2^Z the Boolean Lattice defined on Z (binary code: top=1*, bottom=0*)
and Z! as infinite permutation group with _two_ generators.
(see http://home.iae.nl/users/benschop/ism.htm )
Then countable <--> 1 generator,
and Cantor's uncountable <--> 2 gen's.
-- NB
Nico Benschop
>
> No, this is very very deep. -- David C. Ullrich
Rather, it's jumping to conclusion.
(taking for obvious what has to be proven;-) -- NB
Nico Benschop
>
> On Tue, 29 May 2001 18:19:08 GMT, Nico Benschop <n.ben...@chello.nl>
> wrote:
> >My point: can it be shown that indeed w! (by permuting a countable
> >length initial listing) produces "UNcountably" many reals on [0,1) ?
> >In other words, interpreting UNcountable not so much as "not countable"
> >(a negative property), but as "ESSENTIALLY more than countable" ...
> >You get my drift? This is my "sequential dimension" approach via
> >infinite group Z! (all permutations of the integers Z), which contains
> >2^Z (all subsets of the integers) ... yet Z! has ONLY 2 generators -
> >which is ESSENTIALLY another ballpark alltogether: "NOT countable"
> >requires more than ONE single generator (re: Peano's N = {+1*} iterated
> >successor function), so at least TWO generators: group Z! containing
> >Boolean Lattice 2^Z (the set of all fixed-points[-sets] of the
> >permutations of Z). Think in closures, with a generating operation,
> >rather than just infinite sets (which really have no 'structure';-)
> >
> >Roughly, I guess this has to do with refusing to be convinced by a
> >single (although constructed) counter example in an infinite context:
>
> Yes, it does have to do with refusing to accept one counterexample.
> Cloth-headedness.
> What's that word Chapman uses? Bollocks, that's it. (I'm not sure
> what a bollock is, but I'm pretty sure I understand what it means
> to say a proof is boolocks.)
>
> >MUCH more can be said in Cantor's diagonal case! Don't you agree?
>
> Of course much more can be said. Any time anyone says _anything_
> there's _always_ much more that can be said. The idea that much
> more _needs_ to be said is nonsense.
Well, is not Cantors diagonal argument the essence of much more,
qua method, then the uncountability of the reals on [0,1) -- like
research on P vs. NP algorithms in complexity theory, for instance.
And re your remark [*]: is it your arrogance ('I know it all' ;-(
or just sheer calcification, that makes your vision so rigid
that you can't see - or be interested - beyond the tip of your nose,
that is: in consequences beyond the specific problem at hand?
After a proof, with absolutely minimal means, of:
1. the reals on [0,1) are UNcountable,
2. there are INfinitely many primes beyond the first k of them
[successor m_k +1 of their product m_k has prime divisor(s) > p_k ]
3. the sum of any two integer p-th powers is NOT a p-th power for p>2,
4. the group of units G1 mod 2^k (for k>2) is not cyclic.
...
Then what is so 'stupid' or 'bollocks' in asking, or be interested in:
1' What causes this UNcountability, is it the way they are generated?
2' You skipped over all primes between p_k and m_k : also m_k -1
has the property that m_k +1 has, to conclude the proof. Moreover,
the sets of prime divisors of m_k -1 and of m_k +1 are disjoint,
possibly providing more info about the primes in general
( mod m_k it follows that each even residue is the sum of two units,
that is: in the units group G1, which contains at least all
primes between p_k and m_k )
3' So what DO all these pairsums of p-th powers cover?
It turns out they cover half the units group G1 mod p^k -
for any sufficiently large precision k at base p, prime p>2, and:
each residue mod p^k is the sum of at most 4 p-th power residues.
4' But how many and which generators DOES G1 mod 2^k require?
Well, two suffice, and can be taken {-1,3} for any k>2.
see http://arxiv.org/abs/math.GM/0105029 : 3 is easily shown to be
a semi-primitive root of unity, with some practical engineering
consequences (logarithmic binary code, dual base 2 and 3).
Such results come from asking _around_ a topic, not just being stuck
at it. Hence, rather than nit-picking about a minimal proof re:
Cantor's diagonal argument and uncountability,
Euclid's infinity of primes, and
Fermat's anti-closure of powersums,
look around and see what constructive clues follow from them...
> > (would this have anything to do with my countryman Brouwer?-)
>
> Nothing as far as I can see.
> Did you have anything specific in mind? -- David C. Ullrich
Feast your (hopefully not completely calcified;-) braincells on this
suggestion... lest you meet Chapman, dennis-feldmann, Kabatoff, Quet,
and a few others in my killfile -- as totally useless for any
constructive discussion... (which is a symmetric relation, as far
as I'm concerned, no problem with that - called "plonk" ?-)
Robin Chapman
>
>
>
David C. Ullrich
I didn't say there was anything stupid about that. It's _very_ clear
from what I wrote that what I was calling stupid, and will again if
you insist, is "refusing to be convinced by a single (although
constructed) counter example".
>2' You skipped over all primes between p_k and m_k : also m_k -1
> has the property that m_k +1 has, to conclude the proof. Moreover,
> the sets of prime divisors of m_k -1 and of m_k +1 are disjoint,
> possibly providing more info about the primes in general
> ( mod m_k it follows that each even residue is the sum of two units,
> that is: in the units group G1, which contains at least all
> primes between p_k and m_k )
>
>3' So what DO all these pairsums of p-th powers cover?
> It turns out they cover half the units group G1 mod p^k -
> for any sufficiently large precision k at base p, prime p>2, and:
> each residue mod p^k is the sum of at most 4 p-th power residues.
>
>4' But how many and which generators DOES G1 mod 2^k require?
> Well, two suffice, and can be taken {-1,3} for any k>2.
> see http://arxiv.org/abs/math.GM/0105029 : 3 is easily shown to be
> a semi-primitive root of unity, with some practical engineering
> consequences (logarithmic binary code, dual base 2 and 3).
>
>Such results come from asking _around_ a topic, not just being stuck
>at it. Hence, rather than nit-picking about a minimal proof re:
> Cantor's diagonal argument and uncountability,
I wasn't nit-picking about this proof, I was talking about people
who insist that this proof is _wrong_. Like the web site you referred
to. Or like someone who would refuse to be convinced by a single
counterexample.
> Euclid's infinity of primes, and
> Fermat's anti-closure of powersums,
>look around and see what constructive clues follow from them...
>
>> > (would this have anything to do with my countryman Brouwer?-)
>>
>> Nothing as far as I can see.
>> Did you have anything specific in mind? -- David C. Ullrich
>
>Feast your (hopefully not completely calcified;-) braincells on this
>suggestion...
??? _What_ suggestion? You seemed to be hinting that something
somewhere in this thread had something to do with intuitionism
or some such thing. I don't see what any of it has to do with
anything like that. So I asked what you had in mind - I wish
you'd _explain_ what any of this has to do with Brouwer.
> lest you meet Chapman, dennis-feldmann, Kabatoff, Quet,
>and a few others in my killfile -- as totally useless for any
>constructive discussion... (which is a symmetric relation, as far
>as I'm concerned, no problem with that - called "plonk" ?-)
I really can't say how terrifying I find the idea that you
might killfile me. Well, I _could_ say, but I feel I shouldn't.
But I don't see how this explains what any of this has to
do with Brouwer.
>> >-- NB -- http://home.iae.nl/users/benschop/cantor.htm
>> > http://home.iae.nl/users/benschop/seqdim.htm
>> >
>> >---- One is Always Halfway Anyway --- AHA
>> >xxxxxxxxxxxxxxxxxxx1.1xxxxxxxxxxxxxxxxxxx
David C. Ullrich
David C. Ullrich
wrote:
>"David C. Ullrich" wrote:
>>
[...]
>> >
>> >Lastly, shouldn't ANY countable list of real numbers omit
>> >uncountably many (in fact, more than this -- co-countably many)
>> >real numbers? Isn't this trivial? - Dave L. Renfro
>>
>> No, this is very very deep. -- David C. Ullrich
>
>Rather, it's jumping to conclusion.
> (taking for obvious what has to be proven;-) -- NB
No, he said "trivial". It does need to be proven, but the
proof is trivial:
That _one_ counterexample _proves_ that the reals are
uncountable. And once we know that the reals are
uncountable it is _trivial_ that there are uncountably
many reals not on our list - this is clear _without_
any further construction, because it is immediate from
the fact that the union of two countable sets is
countable.
John Sullivan
2^(Aleph 0) part of the communication problem? Getting back to
basics, aleph-0 is the 'power' of the set N of natural numbers (1,2,3,...).
The word 'power' means something because there exist sets S such that the
elements of N can be put in a one-to-one correspondence with a proper
subset of S but the elements of an S cannot be put in a one-to-one
correspondence with a proper subset of N. It has been proven that
aleph-1, the power of the set of all subsets of N (by definition) is
greater than the power of N, because the elements of an aleph-0 set can
map into a proper subset of the elements of an aleph-1 set, but not
vice versa. And it has been proven that defining each aleph number as
the power of the set of all subsets of a set of the previous aleph number
power yields an infinite series of aleph numbers, each larger than the
previous.
It has also been proven that aleph-1 cannot be either proven or disproven
to be the same as the power of the continuum, ie. the power of the set
of real numbers. Thus we have Cantorian set theory and non-Cantorian
set theory as equally valid.
Now, what in the hell are you people talking about?
JS
Dave Seaman
John Sullivan <su...@altavista.com> wrote:
>Isn't the use of unnecessarily nonsensical notation, such as
>2^(Aleph 0) part of the communication problem? Getting back to
>basics, aleph-0 is the 'power' of the set N of natural numbers (1,2,3,...).
>The word 'power' means something because there exist sets S such that the
>elements of N can be put in a one-to-one correspondence with a proper
>subset of S but the elements of an S cannot be put in a one-to-one
>correspondence with a proper subset of N. It has been proven that
>aleph-1, the power of the set of all subsets of N (by definition) is
That is not the definition of aleph_1. Aleph_1 is simply the next larger
cardinal number after aleph_0. The conjecture that aleph_1 = 2^(aleph_0) is
called the continuum hypothesis, and is known to be independent of the axioms
of ZFC.
>greater than the power of N, because the elements of an aleph-0 set can
>map into a proper subset of the elements of an aleph-1 set, but not
>vice versa.
>And it has been proven that defining each aleph number as
>the power of the set of all subsets of a set of the previous aleph number
>power yields an infinite series of aleph numbers, each larger than the
>previous.
No, that yields an infinite series of what are called beth numbers. The
statement that there are no alephs other than the beth numbers is called the
generalized continuum hypothesis.
>It has also been proven that aleph-1 cannot be either proven or disproven
>to be the same as the power of the continuum, ie. the power of the set
>of real numbers.
That statement, at least, is correct. However, it is easy to show that the
cardinality of the continuum is 2^(aleph_0).
>Thus we have Cantorian set theory and non-Cantorian
>set theory as equally valid.
There is nothing non-Cantorian about any of this.
>Now, what in the hell are you people talking about?
--
Dave Seaman dse...@purdue.edu
Amnesty International calls for new trial for Mumia Abu-Jamal
<http://www.amnestyusa.org/abolish/reports/mumia/>
Dave L. Renfro
[sci.math Wed, 30 May 2001 19:31:42 GMT]
<http://forum.swarthmore.edu/epigone/sci.math/stantaxfan>
wrote (in part and responding to my earlier post):
>> O-K, I'll bite--what's the definition of (w!)? I know what the
>> ORDINAL number w! is (it's ordinal number w^2), but the only
>> reasonable definition that I know of for the CARDINAL w! is the
>> infinite cardinal product 1*2*3*4*..., which is the cardinal w.
>> Maybe you're thinking of blue jays?
>
> Not that... never mind the disclaimers. Surely a reasonable
> definition for the factorial of a cardinal would be the number
> of permutations of a set with that cardinality.
>
>> Oh, and if you define the cardinal w! by using a specific countable
>> set for w, make sure that your definition is well-defined. That is,
>> your definition outputs the same cardinal number no matter what
>> countable set is used.
Yes, that makes sense. I thought NB meant something along these
lines, but I was working on something else when I wrote that and
wasn't paying a lot of attention. The cardinality of the bijective
functions from a set onto itself doesn't change if you use another
set with the same cardinality.
But (for NB), it's trivial that the set of real numbers not
included in a countable list has co-countable complement -- remove
a countable set from an uncountable set and you're left with a
set whose complement is countable. Note that this is a stronger
result than you're claiming, since every co-countable set of
real numbers is uncountable and there are a lot of uncountable
sets of real numbers that are not co-countable. [Examples of the
latter: any interval of real numbers with positive and finite
length, the Cantor set, the positive real numbers, etc.]
Moreover (for NB), if your claim is that you're not just showing
the existence of such numbers, but you're actually claiming to
show some sort of constructive method of obtaining them, then I'm
interested in what your meaning of "constructive" is. Also, are
you saying that each of the numbers you're getting is constructively
obtained or are you saying that the uncountable set of numbers itself
is constructively obtained?
Nonetheless (for everyone), there are a couple of interesting SF
stories by Stephen Baxter that have some relevance to this:
"The Godel Sunflowers" (Interzone 55, January 1992) and "The Logic
Pool" (Isaac Asimov's Science Fiction Magazine, June 1994).
[Both appear in his book "Vacuum Diagrams" --
<http://www.scifi.com/sfw/issue94/books.html>]
"The Logic Pool", for instance, is about AI life that is sort of
like the meme-minds in Gregory Benford's "Foundations Fear".
I think this means some sort of intelligence whose existence
arises from abstract ideas, like beauty and morality (but maybe
more complex than these). The ideas evolve and become modified
over time, analogous to ordinary intelligence (biological and
machine) that arise and evolve from matter/energy things. Anyway,
Baxter's "life" consists of a tree--in the mathematical sense--of
complex and evolving axiomatic theories. Each branch grows by
continuously adding new axioms obtained by using Godel's technique
to obtain statements that are true but not provable in the theory
generated by the set of all statements preceeding (in the
lattice-order sense) the axiom being added.
Really weird stuff.
For more about Baxter, see
<http://www.sam.math.ethz.ch/~pkeller/BAXTER/Interview.html>
<http://www.scifi.com/sfw/issue27/interview.html>
"I suspect the great alien novel is yet to be written:
something totally inhuman, non-specific consciousness,
maybe a different kind of math (able to handle infinity,
for instance, but not simple arithmetic), a totally
alien sensorium... I'll keep trying."
By the way, Alex Kasman (College of Charleston, SC) put out a
request in sci.math on January 31, 2000
<http://forum.swarthmore.edu/epigone/sci.math/flermmordar>
for literature (fiction) having mathematical themes for a web
page he was going to make. I just glanced through it to see
how it's coming along, and it's a pretty good collection now:
<http://math.cofc.edu/faculty/kasman/MATHFICT/Default.html>
Dave L. Renfro
Ross A. Finlayson
Dave Seaman wrote:
> That statement, at least, is correct. However, it is easy to show that the
> cardinality of the continuum is 2^(aleph_0).
>
How is that? It would have to imply a mapping from the powerset of the naturals
to the reals.
Ross
--
Ross A. Finlayson
Finlayson Consulting
Dave L. Renfro
[sci.math 31 May 01 13:30:39 -0400 (EDT)]
<http://forum.swarthmore.edu/epigone/sci.math/stantaxfan>
wrote (in part):
> "The Logic Pool", for instance, is about AI life that is sort of
> like the meme-minds in Gregory Benford's "Foundations Fear".
> I think this means some sort of intelligence whose existence
> arises from abstract ideas, like beauty and morality (but maybe
> more complex than these). The ideas evolve and become modified
> over time, analogous to ordinary intelligence (biological and
> machine) that arise and evolve from matter/energy things. Anyway,
> Baxter's "life" consists of a tree--in the mathematical sense--of
> complex and evolving axiomatic theories. Each branch grows by
> continuously adding new axioms obtained by using Godel's technique
> to obtain statements that are true but not provable in the theory
> generated by the set of all statements preceeding (in the
> lattice-order sense) the axiom being added.
Hummm... I could have said this better. I seem to be confusing
what causes mental states with the mental states themselves.
Here's a better description --->>>
"The Logic Pool", for instance, deals with an intelligence that is
similar to the meme-minds in Gregory Benford's "Foundations Fear".
Meme-mind -- I think this means some sort of intelligence whose
existence arises from abstract ideas, like beauty and morality
(but maybe more complex than these). As the ideas evolve and
become modified over time, the underlying meme-mind achieves
consciousness analogous to ordinary intelligences (biological and
machine) that achieve consciousness when certain matter/energy
things evolve and become modified over time (e.g. consciousness
from a human brain, consciousness from a Pentium 24 chip, etc.).
Anyway, Baxter's lifeform is a tree--in the mathematical
sense--of complex and evolving axiomatic theories. Each branch
grows by continuously adding new axioms obtained by using Godel's
technique to obtain statements that are true but not provable in
the theory generated by the set of all statements preceding (in the
lattice-order sense) the axiom being added.
Dave L. Renfro
Dave Seaman
Ross A. Finlayson <rfinl...@hot.rr.com> wrote:
>Dave Seaman wrote:
>> That statement, at least, is correct. However, it is easy to show that the
>> cardinality of the continuum is 2^(aleph_0).
>How is that? It would have to imply a mapping from the powerset of the naturals
>to the reals.
For anyone who thinks there is a bijection between any two infinite sets,
that would seem to be trivial.
For the rest of us, it's an easy consequence of the Cantor-Bernstein
theorem. The members of P(N) can be represented as infinite binary
strings, and there is a bijection between the real numbers and a proper
subset of the binary strings (namely, the ones that do not terminate in
all zeroes). There is also a natural mapping between the set of all
binary strings and a proper subset of the reals (namely, the Cantor set).
The C-B theorem says that if you have an injection f: A->B and also an
injection g: B->A, then there is a bijection between A and B. Done.
John Sullivan
And the continuum hypothesis is as I said it was. My source is this book
which was one of my math books in college:
Finite Mathematical Structures, by Kemeny, Merkil, Snell and Thompson,
Prentice-Hall, 1959, Chapter 2, sections 12 and 13, Cardinal Numbers and
The Hierarchy of Cardinals, page 102 and following.
JS
a...@seaman.cc.purdue.edu (Dave Seaman) wrote in message news:<9f61ma$j...@seaman.cc.purdue.edu>...
Jesse F. Hughes
>
> For anyone who thinks there is a bijection between any two infinite sets,
> that would seem to be trivial.
>
> For the rest of us, it's an easy consequence of the Cantor-Bernstein
> theorem. The members of P(N) can be represented as infinite binary
> strings, and there is a bijection between the real numbers and a proper
> subset of the binary strings (namely, the ones that do not terminate in
> all zeroes). There is also a natural mapping between the set of all
> binary strings and a proper subset of the reals (namely, the Cantor set).
>
> The C-B theorem says that if you have an injection f: A->B and also an
> injection g: B->A, then there is a bijection between A and B. Done.
C-B requires the axiom of choice, right? Any idea whether
|R| = |P(N)| really depends on choice?
--
"And the logical extension of free and open-source software in the
realm of sex would certainly include publicly shared sex at a sex
party,... queer sexuality and... non-proprietary sexual affection."
Annalee Newitz writing in Salon.com
Dave Seaman
John Sullivan <su...@altavista.com> wrote:
>I'm sorry, but you're wrong. I gave the correct definition of aleph-1.
>And the continuum hypothesis is as I said it was. My source is this book
>which was one of my math books in college:
You did not mention the words "continuum hypothesis" at all. If you are
referring to your statement
>> >It has also been proven that aleph-1 cannot be either proven or disproven
>> >to be the same as the power of the continuum, ie. the power of the set
>> >of real numbers.
then your version of the continuum hypothesis is correct only if aleph_1
is taken to mean what I said, not what you said.
>Finite Mathematical Structures, by Kemeny, Merkil, Snell and Thompson,
>Prentice-Hall, 1959, Chapter 2, sections 12 and 13, Cardinal Numbers and
>The Hierarchy of Cardinals, page 102 and following.
Either you are misquoting the book, or Kemeny et. al. are wrong. I quote
from the master himself. Georg Cantor: Contributions to the Founding of
the Theory of Transfinite Numbers. (English translation by Philip E.B.
Jourdain. Dover Books.)
(p. 96):
If we denote the power of the linear continuum X [...] by o,
we easily see that it may be represented by, amongst others,
the formula:
o = 2^(aleph_0)
(p. 109):
Out of aleph_0 proceeds, by a definite law, the next greater
cardinal number aleph_1, out of this by the same law the next
greater aleph_2, and so on.
(p. 160)
The second number-class Z(aleph_0) is the totality {a} of
ordinal types a of well-ordered aggregates of the cardinal
number aleph_0.
(p. 169, heading of section 16):
The Power of the Second Number-Class is equal to the Second
Greatest Transfinite Cardinal Number Aleph-One.
In other words, aleph_1 is the cardinality of the set of all countably
infinite ordinals, which happens to be the same as the cardinality of the
set of all countable ordinals (including the finite ones). The continuum
hypothesis is the statement that the cardinality of the countable
ordinals is the same as the cardinality of the real numbers.
Dave Seaman
Jesse F. Hughes <je...@andrew.cmu.edu> wrote:
>C-B requires the axiom of choice, right? Any idea whether
>|R| = |P(N)| really depends on choice?
No. Cantor's proof depended on the assumption that every set could be
well-ordered, but Schroeder and Bernstein independently showed how to
prove the result without this assumption.
Nico Benschop
>
> [...]
> Yes, that makes sense. I thought NB meant something along these
> lines, but I was working on something else when I wrote that and
> wasn't paying a lot of attention. The cardinality of the bijective
> functions from a set onto itself doesn't change if you use another
> set with the same cardinality.
>
> But (for NB), it's trivial that the set of real numbers not
> included in a countable list has co-countable complement -- remove
> a countable set from an uncountable set and you're left with a
> set whose complement is countable. Note that this is a stronger
> result than you're claiming, since every co-countable set of
> real numbers is uncountable and there are a lot of uncountable
> sets of real numbers that are not co-countable. [Examples of the
> latter: any interval of real numbers with positive and finite
> length, the Cantor set, the positive real numbers, etc.]
Thanks, I'm not an expert, but I'll look into the co-countable concept
(given time;-). What I am interested in is a different look at the
various types of infinity, that is: from a generative point of view,
which I then (possibly confusingly) call 'constructive'. At least:
the condition I make for sensibly talking about & analysing & comparing
these infinities is this: that there is an operation of which I know
the algebraic composition rule properties (briefly 'syntax') - under
which operation any pair of elements in such infinity produces again
an element in such 'set', which latter I prefer to call a 'system' or
'closure'. This, if i'm not mistaken, is one of the basic concepts in
mathematics (and especially necessary for infinite stuff, since listing
all elements and their relations cq. structure would not be possible;-)
A further restriction that is particularly intersting for me, doing
industrial research on digital system (VLSI) design methods, is that
such operation is at least associative. For modelling deterministic
digital behaviour of hard- and software: the internal state model -
Finite State Machines (FSM) which recently I consider to have an
infinite number of states, namely stateset Z (the integers) -->
yielding Integer State Machines (ISM) , as a more 'sympathetic' model
with a nice algenra (function composition = associtive = semigroups)
than the known Turing Machines (with infinite tape, but no clear
separation between data and input/output sequencies (it's all on tape;-(
This context (infinite semigroups, as closures of ISM's) may explain
my answer to your next question about what I mean with 'constructive'.
> Moreover (for NB), if your claim is that you're not just showing
> the existence of such numbers, but you're actually claiming to
> show some sort of constructive method of obtaining them, then
> I'm interested in what your meaning of "constructive" is. ...[*]
> Also, are you saying that each of the numbers you're getting is
> constructively obtained or are you saying that the uncountable set of
> numbers itself is constructively obtained? [&] -- Dave L. Renfro
On[&]: I can't answer that, since I don't understand the question
(what is the difference between constructing 'each element'
and constructing their 'set' ?-) All I'm interested in is
a precise description of the (potential) infinite set of el's,
like Peano's successor function to h=generate N = 0{+1}*
that is: start with '0' and add 1 repeatedly (ad infinitum).
(If that is not precise enough, don't worry: it _is_ for me;-)
Re[*]: Referring to my explanation above, of the context I'm working in,
the only infinity that I take for granted, as 'base-set' of states in
a state machine approach, is the set of integers Z, under addition(+).
Note that I do not take Peano's naturals N(+) (positive integers)
because that is not a group (complement -n does not exist for n in N).
It is known that each finite symmetric group G_n of all n! permutations
of n states requires just TWO generators. While N(+) with Peano has
just ONE generator (I call it a 'linear' structure), and taking the
complenent of any n in Z for granted, Z(+) also has just ONE generator,
namely '1'.
Notice that also the infinite group Z! of all permutations of Z
has only TWO generators (see http://home.iae.nl/users/benschop/ism.htm
with Ilias Kastanas's analysis; I at first thought that 3 were required)
Now widening our view of Cantor's powerset 2^N to 2^Z is not much of
a jump, is it? Both N and Z are countable sets (I see them as a one-
sided and a two-sided linear infinity, if you get my drift;-)
However, the only operations on the subsets of Z that are relevant,
are intersection and union (Boolean Lattice), which although associative
are both idempotent: not very useful as generating device...
But notice that Z! "contains" 2^Z, by considering the set of
fixed-points
of each permutation of Z. Such fixedpointset, which can be any subset
of Z, defines a maximal (infinite) stabilizer group, as subgroup of Z!
Hence Z! contains 2^Z, while Z! has only 2 generators, e.g the +1
successor function with its inverse -1, cyclic permuting all of Z
with no fixed points, and the 2-cycle: permute 0 <--> 1 while fixing
all other states = integers <> {0,1}.
Notice that by approaching 2^Z "from above" via Z! we DO have a handle
on its generation ('finitely generated: just 2 generators;-) and its
representation, including intersect & union operation of 2^Z as Boolean
Lattice, but also the additional 'sequencing' operation of permutations
which ia associative (be they finite or infinite: 2 generators
suffice;-)
What is ambiguous about defining the cardinal w! (omega factorial)
as the cardinality of Z! -- which must necessarily be at least that
of 2^Z since the latter system is properly contained in group Z!
It may well be that these two cardinalities: 2^w and w! are equal,
but certainly this w! cannot possibly be smaller that 2^w, can it?
My final suggestion then is: do NOT look at the "size" (cardinality)
of infinite sets, but at the minimal number of GENERATORS ("seqdim")
of infinite SYSTEMS (_with_ generating rule;-). You see?
We then do NOT compare apples with oranges (integers with reals)
because they are of DIFFERENT TYPE, namely members of ONE- resp. TWO-
generator systems, which should NOT be compared in 'size', at least
not in the infinite case...
And then there still is Z^Z : the system (under concatenation) of all
transformations of Z (functions f: Z --> Z), with ONLY 3 GENERATORS:
Z^Z = {+1, swap2, merge2}* ... Only then the real fun begins!!!
-- NB -- http://home.iae.nl/users/benschop
http://home.iae.nl/users/benschop/seqdim.htm
http://arXiv.org/abs/math.GM/0103112
(On semigroups, and the 5 basic state machines)
[towards a Digital Network Theory = Semigroup structure theory]
[as coupled network of the 5 basic types of sequential beviour]
[Balancing Syntax & Semantics : Soft- & Hardware : Algebra &
Represent'n]
Robin Chapman
>of n states requires just TWO generators. While N(+) with Peano has
>just ONE generator (I call it a 'linear' structure), and taking the
>complenent of any n in Z for granted, Z(+) also has just ONE generator,
>namely '1'.
> Notice that also the infinite group Z! of all permutations of Z
>has only TWO generators (see http://home.iae.nl/users/benschop/ism.htm
>with Ilias Kastanas's analysis; I at first thought that 3 were required)
No. Two (or finitely many, or countably many) permutations will
only generate a countable subgroup of the permutations of Z.
To get all this group requires uncountably many generators.
>
>But notice that Z! "contains" 2^Z, by considering the set of
>fixed-points
>of each permutation of Z. Such fixedpointset, which can be any subset
>of Z, defines a maximal (infinite) stabilizer group, as subgroup of Z!
A funny use of the word "contains". Anyway not every subset of Z
is the fixed point set of a permutation of Z.
>
>What is ambiguous about defining the cardinal w! (omega factorial)
>as the cardinality of Z! -- which must necessarily be at least that
> of 2^Z since the latter system is properly contained in group Z!
>It may well be that these two cardinalities: 2^w and w! are equal,
>but certainly this w! cannot possibly be smaller that 2^w, can it?
The cardinalities of the sym(Z) and P(Z) are the same.
David C. Ullrich
wrote:
>Isn't the use of unnecessarily nonsensical notation, such as
>2^(Aleph 0) part of the communication problem?
I don't see what's nonsensical about the notation - as long as
we know what it means then that's what it means. I also don't
see what this has to do with that proof, where this notation
does not arise. And I _really_ don't see how _any_ notation,
nonsensical or not, is a reasonable explanation for people
saying that mathematicians are all wrong, when in fact those
people simply don't follow simple logic. The confusion
has nothing to do with the notation for cardinals, the
confusion has to with not realizing that
Suppose X has property P.
[argument]
Then X does not have property Q
is a valid proof that there is no X with both properties
P and Q. When someone says that a proof as above is
fatally flawed, and explains that the flaw is
Suppose X has property P.
[argument]
Then X does not have property Q. But wait, there's
this other value X' that _does_ have property Q.
that person is being cloth-headed.
> Getting back to
>basics, aleph-0 is the 'power' of the set N of natural numbers (1,2,3,...).
>The word 'power' means something because there exist sets S such that the
>elements of N can be put in a one-to-one correspondence with a proper
>subset of S but the elements of an S cannot be put in a one-to-one
>correspondence with a proper subset of N. It has been proven that
>aleph-1, the power of the set of all subsets of N (by definition)
No, that is _not_ the definition of aleph_1. Aleph_1 is the first
uncountable ordinal - you're talking about 2^Alpeh_0 here, which
_may_ or _may_ _not_ be the same as Aleph_1.
Good of you to try to clarify things. Jeez...
> is
>greater than the power of N, because the elements of an aleph-0 set can
>map into a proper subset of the elements of an aleph-1 set, but not
>vice versa. And it has been proven that defining each aleph number as
>the power of the set of all subsets of a set of the previous aleph number
>power yields an infinite series of aleph numbers, each larger than the
>previous.
This is _not_ the definition of those alephs.
>It has also been proven that aleph-1 cannot be either proven or disproven
>to be the same as the power of the continuum, ie. the power of the set
>of real numbers.
It is true that aleph_1 cannot be proven to be or not to be the same
as c, the power of the continuum. But you're very confused about all
this, because aleph_1 is not what you think it is. It is very
easy to prove that c = 2^aleph_0, where 2^aleph_0 is by definition
the cardinality of the power set of the integers.
> Thus we have Cantorian set theory and non-Cantorian
>set theory as equally valid.
Huh???????? There's nothing "non-Cantorian" about anything
you've said (the true things or the false ones either.)
>Now, what in the hell are you people talking about?
Good one.
>JS
David C. Ullrich
<rfinl...@hot.rr.com> wrote:
>
>
>Dave Seaman wrote:
>
>> That statement, at least, is correct. However, it is easy to show that the
>> cardinality of the continuum is 2^(aleph_0).
>>
>
>How is that? It would have to imply a mapping from the powerset of the naturals
>to the reals.
Well duh. Yes, it would have to imply that. In fact it _does_ imply
that. This is no problem because there _is_ a bijection between the
powerset of the naturals and the reals.
Why don't you _learn_ _some_ _mathematics_ instead of making a fool
of yourself every day?
>Ross
>--
>Ross A. Finlayson
>Finlayson Consulting
>
>
David C. Ullrich
David C. Ullrich
wrote:
>I'm sorry, but you're wrong. I gave the correct definition of aleph-1.
>And the continuum hypothesis is as I said it was. My source is this book
>which was one of my math books in college:
>
>Finite Mathematical Structures, by Kemeny, Merkil, Snell and Thompson,
>Prentice-Hall, 1959, Chapter 2, sections 12 and 13, Cardinal Numbers and
>The Hierarchy of Cardinals, page 102 and following.
Give an _exact_ _verbatim_ quote of the definition of aleph_1 in this
book (so we can tell whether you're misquoting it or the book is
wrong.)
>JS
Ross A. Finlayson
David C. Ullrich wrote:
> On Thu, 31 May 2001 20:04:16 GMT, "Ross A. Finlayson"
> <rfinl...@hot.rr.com> wrote:
>
> >
> >
> >Dave Seaman wrote:
> >
> >> That statement, at least, is correct. However, it is easy to show that the
> >> cardinality of the continuum is 2^(aleph_0).
> >>
> >
> >How is that? It would have to imply a mapping from the powerset of the naturals
> >to the reals.
>
> Well duh. Yes, it would have to imply that. In fact it _does_ imply
> that. This is no problem because there _is_ a bijection between the
> powerset of the naturals and the reals.
>
> Why don't you _learn_ _some_ _mathematics_ instead of making a fool
> of yourself every day?
>
It was a leading question, where I had an answer.
What if there were somewhat less than |N|-1 rationals in the unit interval, with
2^|N| reals in that interval, with the complement (equivalence class) of the
rationals in that unit interval being the irrationals? This is in consideration of
x! / (x-2)!, the count of permutations of a set x into composites of two elements, or
x(x-1), where some of the composites evaluate to the same value as other ones thus
being non-unique.
In this consideration of bit strings as the mapping between P(N) and R, it is so that
for any bit string of base b (here 2) of any given length p that the count of
in-place permutations is b^p. Place the radix at the beginning of the string. One
thing to note is that not all of the rationals end with .000..., and that for any of
the finite elements N, that all strings of length zero to n are rational, for n in
N. For no n in N is the number irrational, there showing 2^n rationals.
Now then, with this bit stream, remove the radix, and the number is always an
integer, in N.
Virgil
Dave Seaman
Ross A. Finlayson <rfinl...@hot.rr.com> wrote:
>What if there were somewhat less than |N|-1 rationals in the unit interval, with
>2^|N| reals in that interval, with the complement (equivalence class) of the
>rationals in that unit interval being the irrationals? This is in consideration of
>x! / (x-2)!, the count of permutations of a set x into composites of two elements, or
>x(x-1), where some of the composites evaluate to the same value as other ones thus
>being non-unique.
<...snip...>
This is not mathematics.
This is typing.
Did you use a prose generator, seeded with a few mathematical terms, to
produce this?
Fred Galvin
> >===== Original Message From Nico Benschop <n.ben...@chello.nl> =====
> >
> >It is known that each finite symmetric group G_n of all n! permutations
> >of n states requires just TWO generators. While N(+) with Peano has
> >just ONE generator (I call it a 'linear' structure), and taking the
> >complenent of any n in Z for granted, Z(+) also has just ONE generator,
> >namely '1'.
> > Notice that also the infinite group Z! of all permutations of Z
> >has only TWO generators (see http://home.iae.nl/users/benschop/ism.htm
> >with Ilias Kastanas's analysis; I at first thought that 3 were required)
>
> No. Two (or finitely many, or countably many) permutations will
> only generate a countable subgroup of the permutations of Z.
And, conversely, every countable subgroup of Z! is contained in a
2-generator subgroup. I think I may have mentioned that fact in an
earlier thread; if Mr. Benschop is one of those who hold that
2^(Aleph_0) = Aleph_0 (I'm not very good at keeping track of the
belief systems of all the people who contribute to sci.math), that
would naturally lead him to conclude that Z! itself is generated by
two permutations. Of course, if you *could* find 3 permutations that
generate Z!, you could replace them with 2 permutations.
David C. Ullrich
<rfinl...@hot.rr.com> wrote:
>
>
>David C. Ullrich wrote:
>
>> On Thu, 31 May 2001 20:04:16 GMT, "Ross A. Finlayson"
>> <rfinl...@hot.rr.com> wrote:
>>
>> >
>> >
>> >Dave Seaman wrote:
>> >
>> >> That statement, at least, is correct. However, it is easy to show that the
>> >> cardinality of the continuum is 2^(aleph_0).
>> >>
>> >
>> >How is that? It would have to imply a mapping from the powerset of the naturals
>> >to the reals.
>>
>> Well duh. Yes, it would have to imply that. In fact it _does_ imply
>> that. This is no problem because there _is_ a bijection between the
>> powerset of the naturals and the reals.
>>
>> Why don't you _learn_ _some_ _mathematics_ instead of making a fool
>> of yourself every day?
>>
>
>It was a leading question, where I had an answer.
>
>What if there were somewhat less than |N|-1 rationals in the unit interval,
As I imagine people have tried to point out before, subtraction of
ordinals and cardinals is undefined. There's no such thing as "|N|-1".
>with
>2^|N| reals in that interval,
There are in fact |N| rationals in the unit interval, and there
are in fact 2|N| reals in that interval. I'm not sure what to
make of the "what if".
>with the complement (equivalence class)
Using words you don't understand even a little bit again.
"Equivalence class" makes no sense in this context.
>f the
>rationals in that unit interval being the irrationals?
Again, the complement of the rationals in the interval
_is_ the set of irrationals in the interval - the "what
is" remains puzzling...
Oh - that's the end of the question. What do you mean,
"What if?" Various things you said don't quite make sense,
but if a person tries to figure out what you mean they're
all simply _true_. What is your question, exactly? _What_
"what if? What if _what_?
(Suppose someone asks you "What if 2 + 2 were equal to 4?"
How would you reply to that?)
>This is in consideration of
>x! / (x-2)!, the count of permutations of a set x into composites of two elements, or
>x(x-1), where some of the composites evaluate to the same value as other ones thus
>being non-unique.
Greeble fooclumnish, or perhaps.
>In this consideration of bit strings as the mapping between P(N) and R, it is so that
>for any bit string of base b (here 2) of any given length p that the count of
>in-place permutations is b^p. Place the radix at the beginning of the string. One
>thing to note is that not all of the rationals end with .000..., and that for any of
>the finite elements N, that all strings of length zero to n are rational, for n in
>N. For no n in N is the number irrational, there showing 2^n rationals.
>
>Now then, with this bit stream, remove the radix, and the number is always an
>integer, in N.
Om.
>Ross
>--
>Ross A. Finlayson
>Finlayson Consulting
>
>
David C. Ullrich
John Sullivan
> On 31 May 2001 14:32:49 -0700, su...@altavista.com (John Sullivan)
> wrote:
> >I'm sorry, but you're wrong. I gave the correct definition of aleph-1.
> >And the continuum hypothesis is as I said it was. My source is this book
> >which was one of my math books in college:
> >Finite Mathematical Structures, by Kemeny, Merkil, Snell and Thompson,
> >Prentice-Hall, 1959, Chapter 2, sections 12 and 13, Cardinal Numbers and
> >The Hierarchy of Cardinals, page 102 and following.
> Give an _exact_ _verbatim_ quote of the definition of aleph_1 in this
> book (so we can tell whether you're misquoting it or the book is
> wrong.
The book defines aleph-0 as the cardinality of the
set of positive integers. Then it says,
"The next lowest cardinal is called aleph-1, the
next aleph-2, etc."
But the book does not say why it claims that there
exists a "next lowest cardinal". What is does do
is show that the cardinality of the set of all
subsets of an aleph-0 set is greater than aleph-0.
And it shows that the cardinality of the set of all
subsets of a set of cardinality greater than aleph-0
is greater than the cardinality of that set, etc,
leading to an infinite series of cardinalities,
each greater than the previous.
I naturally assumed that the cardinalities of
these 'sets of all subsets' were the definitions of
aleph-1, aleph-2, etc.
If I was not to assume that, then how could I accept the
proposition that there is a 'next lowest cardinal'
greater than aleph-0? In other areas, I know that there
is no 'next lowest'. For example there is no next
lowest rational number compared to 1/2, and there is
no next lowest irrational number compared to pi,
in fact no 'adjacent' real numbers at all, so why should
I accept that there is a next lowest cardinality
after aleph-0? The way I accepted it was to assume
that the aforementioned progression of cardinalities of
sets of all subsets was meant to define the aleph
numbers.
Apparently I was wrong to assume that, but I am left
wondering how to accept that there exists a 'next lowest'
aleph number after aleph-0. Remember, I'm not now
allowed to make use of 'sets of all subsets' in this
effort.
The book goes on to describe the 'continuum hypothesis'
as the conjecture that the cardinality of the set of
real numbers is aleph-1. But I now see that aleph-1
has been defined only as above, and not as I thought.
The book also leaves as an exercize the proof that the
cardinality of the set of all subsets of the positive
integers is the same as the cardinality of the set of
real numbers. I never noticed that exercize before.
Naturally, if that was the continuum hypothesis, as I
thought, it would not have been an exercize, except for
a student named Godel perhaps.
JS
Jesse F. Hughes
> The book defines aleph-0 as the cardinality of the
> set of positive integers. Then it says,
>
> "The next lowest cardinal is called aleph-1, the
> next aleph-2, etc."
>
> But the book does not say why it claims that there
> exists a "next lowest cardinal". What is does do
> is show that the cardinality of the set of all
> subsets of an aleph-0 set is greater than aleph-0.
> And it shows that the cardinality of the set of all
> subsets of a set of cardinality greater than aleph-0
> is greater than the cardinality of that set, etc,
> leading to an infinite series of cardinalities,
> each greater than the previous.
>
> I naturally assumed that the cardinalities of
> these 'sets of all subsets' were the definitions of
> aleph-1, aleph-2, etc.
>
> If I was not to assume that, then how could I accept the
> proposition that there is a 'next lowest cardinal'
> greater than aleph-0?
Cardinals are a subset of the ordinals. The ordinals are
well-founded, meaning that any non-empty collection of ordinals has a
least element. Hence, the collection of all ordinals which are
cardinals greater than aleph-0 has a least element, which we call
aleph-1.
I hope the book mentioned well-foundedness (or transfinite induction)
in passing.
--
"[I]t's good for the economy to charge for intellectual property, so
open source software cannot be good, while Microsoft is the most
far-thinking company around and is doing it all for the good of the
public." -- Linus Torvalds paraphrases Microsoft VP Craig Mundie
Dave Seaman
John Sullivan <su...@altavista.com> wrote:
>The book defines aleph-0 as the cardinality of the
>set of positive integers. Then it says,
>"The next lowest cardinal is called aleph-1, the
>next aleph-2, etc."
>But the book does not say why it claims that there
>exists a "next lowest cardinal".
Just as the set of all finite ordinals is not a finite set, so too is the
set of all countable ordinals not a countable set. We say w_0 is the
first transfinite ordinal, since each member of w_0 is finite, but there
are infinitely many such members. We say w_1 is the first uncountable
ordinal, since each member of w_1 is countable, but there are uncountably
many of them. After all, if w_1 were countable, it would have to be a
member of itself, which is forbidden in ZFC.
>What is does do
>is show that the cardinality of the set of all
>subsets of an aleph-0 set is greater than aleph-0.
>And it shows that the cardinality of the set of all
>subsets of a set of cardinality greater than aleph-0
>is greater than the cardinality of that set, etc,
>leading to an infinite series of cardinalities,
>each greater than the previous.
>I naturally assumed that the cardinalities of
>these 'sets of all subsets' were the definitions of
>aleph-1, aleph-2, etc.
>If I was not to assume that, then how could I accept the
>proposition that there is a 'next lowest cardinal'
>greater than aleph-0? In other areas, I know that there
>is no 'next lowest'. For example there is no next
>lowest rational number compared to 1/2, and there is
>no next lowest irrational number compared to pi,
>in fact no 'adjacent' real numbers at all, so why should
>I accept that there is a next lowest cardinality
>after aleph-0? The way I accepted it was to assume
>that the aforementioned progression of cardinalities of
>sets of all subsets was meant to define the aleph
>numbers.
The cardinals are often defined to be the initial ordinals; such an
ordinal has a larger cardinality than any of its members. For example,
w_0 and w_1 are both initial ordinals, and therefore are also cardinals.
When we want to think of w_0 as a cardinal, we give it the name aleph_0.
In like manner, aleph_1 is simply w_1, viewed as a cardinal. Thus, the
cardinals are a subclass of the ordinals. Since the ordinals are
well-ordered, it follows that the cardinals are well-ordered. There is a
first uncountable ordinal, and that ordinal is also a cardinal, hence it
is also the first uncountable cardinal.
Robin Chapman
>On Fri, 1 Jun 2001, Robin Chapman wrote:
>
>> >===== Original Message From Nico Benschop <n.ben...@chello.nl> =====
>> >
>> >It is known that each finite symmetric group G_n of all n! permutations
>> >of n states requires just TWO generators. While N(+) with Peano has
>> >just ONE generator (I call it a 'linear' structure), and taking the
>> >complenent of any n in Z for granted, Z(+) also has just ONE generator,
>> >namely '1'.
>> > Notice that also the infinite group Z! of all permutations of Z
>> >has only TWO generators (see http://home.iae.nl/users/benschop/ism.htm
>> >with Ilias Kastanas's analysis; I at first thought that 3 were required)
>>
>> No. Two (or finitely many, or countably many) permutations will
>> only generate a countable subgroup of the permutations of Z.
>
>And, conversely, every countable subgroup of Z! is contained in a
>2-generator subgroup. I think I may have mentioned that fact in an
>earlier thread; if Mr. Benschop is one of those who hold that
>2^(Aleph_0) = Aleph_0 (I'm not very good at keeping track of the
>belief systems of all the people who contribute to sci.math),
It's hard to tell, as his literary style is so opaque, but it
seems to me he has moved away from 2^(aleph_0) = aleph_0 to
2^(aleph_0) > aleph_0. On the other hand he still doesn't
like the Cantor proof very much.
> that
>would naturally lead him to conclude that Z! itself is generated by
>two permutations. Of course, if you *could* find 3 permutations that
>generate Z!, you could replace them with 2 permutations.
But he claims to have two explicit permutations to do this!
From his webpage http://home.iae.nl/users/benschop/cant-ism.htm
we find
"Z! has only 2 generators, e.g the +1
successor function with its inverse -1, cyclic permuting all of Z
with no fixed points, and the 2-cycle: permute 0 <--> 1 while fixing
all other states = integers <> {0,1}."
It seems he believes that Z! is generated by x |-> x+1 and
the transposition (0 1). Perhaps we should dub the (countable) group
these actually generate as the "Benschop group" B and investigate
its properties.
Well, it's really pretty obvious that B is the semidirect product of
the group of permutations of Z with cofinite fixed pont set
and an infinite cyclic group. Indeed B is the set of permutations
sigma such that there exist n and N such that sigma(a) = a + n whenever
|a| > N. So even the negation a |-> -a isn't in there :-)
Dave L. Renfro
<http://forum.swarthmore.edu/epigone/sci.math/stantaxfan>
wrote on 31 May 01 13:30:39 -0400 (EDT) (in part):
> "The Logic Pool", for instance, is about AI life that is sort of
> like the meme-minds in Gregory Benford's "Foundations Fear".
Then Renfro wrote on 31 May 01 15:52:54 -0400 (EDT) (in part):
> Hummm... I could have said this better. I seem to be confusing
> what causes mental states with the mental states themselves.
> Here's a better description --->>>
[snip]
> Anyway, Baxter's lifeform is a tree--in the mathematical
> sense--of complex and evolving axiomatic theories. Each branch
> grows by continuously adding new axioms obtained by using Godel's
> technique to obtain statements that are true but not provable in
> the theory generated by the set of all statements preceding (in the
> lattice-order sense) the axiom being added.
Even what I wrote the second time doesn't make sense. A tree is a
certain type of partially ordered set. In the story, if I understand
it correctly, the elements of the tree are axioms and the chains
in the lattice are axiomatic theories. The tree itself isn't
(necessarily) an axiomatic theory and the elements making up the
tree are not axiomatic theories (except in a degenerate sense,
I suppose -- a set consisting of one axiom is a consistent theory).
At least, this is what I think is going on in Stephen Baxter's story.
**************************************
"The Logic Pool" deals with an intelligence that is similar
to the meme-minds in Gregory Benford's "Foundations Fear".
Meme-mind -- I think this means some sort of intelligence whose
existence arises from abstract ideas, like beauty and morality
(but maybe more complex than these). As the ideas evolve and
become modified over time, the underlying meme-mind achieves
consciousness analogous to ordinary intelligences (biological and
machine) that achieve consciousness when certain matter/energy
things evolve and become modified over time (e.g. consciousness
from a human brain, consciousness from a Pentium 24 chip, etc.).
Anyway, Baxter's lifeform is a tree--in the mathematical sense--that
is a complex and evolving collection of axiomatic theories. Each
chain in the tree is an axiomatic theory and each branch grows by
continuously adding new axioms obtained by using Godel's technique
to obtain statements that are true but not provable in the theory
generated by the set of all statements preceding the axiom being
added (in the lattice-order sense).
**************************************
This sci.math thread is also a tree. If we consider each post to
be an axiom, do any of the maximal chains in this tree form a
consistent theory? (I doubt it.) Is this thread getting complex
enough to have a mind of its own? Maybe when you decide to respond
it's not YOU choosing to respond, it's the thread's MEME-MIND
forcing you to respond. Scary, huh?
Dave L. Renfro
David C. Ullrich
wrote:
That's clear. Probably when you say something on sci.math,
various people say that it's not so, then before insisting
that a certain book agrees with you you should check that
it's actually the book agreeing with you, and not just
your _assumptions_ about what the book means to be saying.
Just a suggestion. You _said_ that the book _says_
that aleph_1=2^aleph_0, and it simply doesn't say that.
>If I was not to assume that, then how could I accept the
>proposition that there is a 'next lowest cardinal'
>greater than aleph-0?
There's no explanation for this given in that book.
It's not actually a book on set theory - when you
see a few optional sections on infinite cardinals
in a book on finite math you should probably assume
that there's more to the story.
(As for why there _is_ a next cardinal after aleph_0,
I imagine that various replies that I see exist
but have not opened yet address this. If not I'll
get back to you on that.)
> In other areas, I know that there
>is no 'next lowest'. For example there is no next
>lowest rational number compared to 1/2, and there is
>no next lowest irrational number compared to pi,
>in fact no 'adjacent' real numbers at all, so why should
>I accept that there is a next lowest cardinality
>after aleph-0? The way I accepted it was to assume
>that the aforementioned progression of cardinalities of
>sets of all subsets was meant to define the aleph
>numbers.
>
>Apparently I was wrong to assume that, but I am left
>wondering how to accept that there exists a 'next lowest'
>aleph number after aleph-0. Remember, I'm not now
>allowed to make use of 'sets of all subsets' in this
>effort.
>
>The book goes on to describe the 'continuum hypothesis'
>as the conjecture that the cardinality of the set of
>real numbers is aleph-1. But I now see that aleph-1
>has been defined only as above, and not as I thought.
And the book also does not say anything about the
fact that c = 2^aleph_0, which it _really_ should.
If you want to learn some set theory try Halmos
"Naive Set Theory". You'll get a much more complete
picture than what's in those sections of that book.
>The book also leaves as an exercize the proof that the
>cardinality of the set of all subsets of the positive
>integers is the same as the cardinality of the set of
>real numbers. I never noticed that exercize before.
Oh - I didn't look at the exercises. If it's at least
in the exercises then it's not quite so bad that
it's not in the text proper.
There's a moral here: Don't skip the exercises -
in advanced math books there is often a lot of
information there.
>Naturally, if that was the continuum hypothesis, as I
>thought, it would not have been an exercize, except for
>a student named Godel perhaps.
>
>JS
David C. Ullrich
David C. Ullrich
I just glanced at those two sections yesterday at the office,
but I don't think it does. (It's just two optional sections
in an undergraduate book on "finite math", not a book on
set theory.)
>"[I]t's good for the economy to charge for intellectual property, so
>open source software cannot be good, while Microsoft is the most
>far-thinking company around and is doing it all for the good of the
>public." -- Linus Torvalds paraphrases Microsoft VP Craig Mundie
David C. Ullrich
Nico Benschop
>
> On Fri, 1 Jun 2001, Robin Chapman wrote:
>
> > >=== Original Message From Nico Benschop <n.ben...@chello.nl> ===
> > >
Interesting... My questions:
1. How does the cardinality of Boolean Lattice 2^Z (Cantor's powerset
of Z) compare with that of Z! (the group of all permutations of Z) ?
Is not card(2^Z) <= card(Z!) ? ... since 2^Z is "contained" in Z!
2. Assuming (correct me if I'm wrong) all binary strings, thus over
{0,1} - of countable length ("to the right of the binary point")
represent
all reals on interval [0,1) (which is an uncountable set;-) ,
how come that all strings over {a,b,c} of countable length, where:
a= +1, b= -1, c= swap(0,1) leaving rest of Z fixed,
does not generate (uncountable;-) Z! = all permutations of Z ...?
(under permutation composition; note that I take -1, complement of
+1,
not for granted, although maybe I should, to have just 2 gen's).
John Sullivan
Thank you for your helpfulness, advice, and the Halmos reference.
-JS
Dave L. Renfro
[sci.math Fri, 01 Jun 2001 12:10:13 +0200]
<http://forum.swarthmore.edu/epigone/sci.math/stantaxfan>
wrote (in part):
> What is ambiguous about defining the cardinal w! (omega factorial)
> as the cardinality of Z! -- which must necessarily be at least that
> of 2^Z since the latter system is properly contained in group Z!
> It may well be that these two cardinalities: 2^w and w! are equal,
> but certainly this w! cannot possibly be smaller that 2^w, can it?
>
> My final suggestion then is: do NOT look at the "size" (cardinality)
> of infinite sets, but at the minimal number of GENERATORS ("seqdim")
> of infinite SYSTEMS (_with_ generating rule;-). You see?
> We then do NOT compare apples with oranges (integers with reals)
> because they are of DIFFERENT TYPE, namely members of ONE- resp.
> TWO-generator systems, which should NOT be compared in 'size',
> at least not in the infinite case...
As you're considering them, 2^w is the same cardinal number as w!.
But I think one of the problems you're having is that you're not
properly isolating your attention to the mathematical properties
you're using. Cardinality is a very rough sort of tool -- it doesn't
have anything to do with the objects you have, their internal
structure (if any), how the objects might or might not be combined,
what color they are, etc. All it tells you, in a generalized sense,
is how many objects you have. Let's say I bring my car to an auto
repair shop and the guy (these shops are almost always run by guys)
asks me what kind of car I have. If I say, "well, it's got
6,396 parts", this conveys almost nothing to him. What if I put
6,396 grains of sand in a cup. Then my car and the cup have the
same number of parts. So is my car like a cup of sand?
Hummm... since my car is a 1986 model with 170,000 miles on it,
this might not be a good example after all . . .
This probably doesn't have much to do with what you're after, but
you (and maybe others) might find this interesting --->>>
Robert Gray, "Georg Cantor and transcendental numbers", American
Mathematical Monthly 101 (1994), 819-832.
This article deals with some ways of obtaining transcendental
numbers via diagonalizations of the algebraic numbers. For instance,
Theorem 3 on page 825 states: "A real number in the interval (0,1)
is transcendental if and only if it is the diagonal number of a
sequence that consists of all the binary representations of
algebraic reals in (0,1)."
Dave L. Renfro
Fred Galvin
> 1. How does the cardinality of Boolean Lattice 2^Z (Cantor's powerset
> of Z) compare with that of Z! (the group of all permutations of Z) ?
> Is not card(2^Z) <= card(Z!) ? ... since 2^Z is "contained" in Z!
Yes, card(2^Z) <= card(Z!), and also card(Z!) <= card(2^Z), whence
card(Z!) = card(2^Z). For the second inequality, a permutation of Z
can be identified with its graph, which is a subset of ZxZ; and there
is a bijection between ZxZ and Z.
> 2. Assuming (correct me if I'm wrong) all binary strings, thus over
> {0,1} - of countable length ("to the right of the binary point")
> represent
> all reals on interval [0,1) (which is an uncountable set;-) ,
> how come that all strings over {a,b,c} of countable length, where:
> a= +1, b= -1, c= swap(0,1) leaving rest of Z fixed,
> does not generate (uncountable;-) Z! = all permutations of Z ...?
> (under permutation composition; note that I take -1, complement of
> +1,
> not for granted, although maybe I should, to have just 2 gen's).
Indeed, the set of all infinite words (strings of length omega) over a
3-letter (or a 2-letter) alphabet has the same cardinality as Z!; and
so it is possible to represent elements of Z! as infinite words over
{a,b,c}; I wouldn't be surprised if it is possible to do so in some
more or less natural way. The words, of course, are just formal
expressions; I don't think there is any sensible way to regard them as
infinite *products* of permutations. For example, what permutation is
the infinite product d = ccc...? And what is cd? Anyway, the standard
meaning of the terminology "the group G is generated by its elements
a, b, c" is that each element of G is equal to some *finite* word in
the generators a, b, c. There is another notion of generation which
may be closer to what you have in mind. A *topological* group is both
a group and a topological space, and the group operations
(multiplication and inversion) are required to be continuous. Now, one
may speak of a *topological* group G being *topologically* generated
by some of its elements. That means that the subgroup of G generated
*algebraically* (finite words only!) by those elements is dense in G,
i.e., each element of G can be *approximated* (in the sense of the
given topology) by finite combinations of the generators. Now, if the
group Z! is topologized in the natural way (the topology of pointwise
convergence), then it is topologically generated by your a, b, c.
Nico Benschop
Thanks Fred, now we're getting somewhere;-) I understand that one
cannot be careful enough with those infinite groups, like Z!
It's just that I compared Cantor's (binary) w x w list, or rather
table (countable set of rows, each of countable length omega 'w')
with a
binary alphabtet B = {0,1} generating all strings B* over B (with that
infamous arithmetic 'real' interpretation r = \sum 2^{-i} with i in N)
-- and the resulting 'uncountable' reals via his diagonal argument --
with the infinite group Z! = A* with two generators A = {add+1, swap2}.
Of course the operation (string concatenation) here means 'permutation
composition' (of permutations defined formally over Z) which must be
defined properly with all e'gars (;-) as a mathematical associative
group operation deserves... Something like you describe above, maybe
in some continuous topologic space, or simply by an algorithm such
as Peano's successor function S(n) = n+1 for n \in N- and using Ilias
Kastanas' type of analysis, re http://home.iae.nl/users/benschop/ism.htm
You restore my faith in the possibility of having a constructive
discussion with a mathematician (requiring learning at both sides;-) ,
which for a while seemed impossible (re: the likes of RC & DU ;-(
I read in the Mac Tutor history site (St.Andrews, Scotland) that
L.E.J. Brouwer (early 20-th century) is considered to 'start' topology
as a discipline (with some famous fixed-point theorem ?-). Also, that
he later disavowed his own contributions to topology, because he changed
his mind, by not accepting the 'excluded middle' (only true/false) in
logic & math-proofs.
This attitude of ignoring proofs by contradiction would only make sense
in an infinite context, I presume... (in a finite context it is
perfectly
logical to accept a proof by contradiction, as a shortcut as it were;-)
Just imagine that Cantor's diagonal procedure would only produce
countably many reals outside any initial listing, so that 'almost all'
reals would be in such list ('almost all' in the infinite context
meaning)
Then the reals on [0,1] would NOT be uncountable, is'nt it? And should
not this impossibility be part of the proof of 'uncountable' ?
David C. Ullrich
wrote:
>I'm sorry, but you're wrong. I gave the correct definition of aleph-1.
>And the continuum hypothesis is as I said it was. My source is this book
>which was one of my math books in college:
>
>Finite Mathematical Structures, by Kemeny, Merkil, Snell and Thompson,
>Prentice-Hall, 1959, Chapter 2, sections 12 and 13, Cardinal Numbers and
>The Hierarchy of Cardinals, page 102 and following.
Having some free time I looked it up. On page 102 it says that aleph_0
is the smallest infinite cardinal and that the next one is called
aleph_1. Which is exactly correct.
Then there is a theorem that proves that 2^alpeph_0 > aleph_0,
which is also exactly correct. But I can't find the spot where
it says that aleph_1 = 2^aleph_0 as you claim. What _page_ is it on?
(It would also be good if you would clarify what's "non-Cantorian"
about anything you said...)
>JS
>
>
>a...@seaman.cc.purdue.edu (Dave Seaman) wrote in message news:<9f61ma$j...@seaman.cc.purdue.edu>...
>> In article <5f6a0413.0105...@posting.google.com>,
>> John Sullivan <su...@altavista.com> wrote:
>> >Isn't the use of unnecessarily nonsensical notation, such as
>> >2^(Aleph 0) part of the communication problem? Getting back to
>> >basics, aleph-0 is the 'power' of the set N of natural numbers (1,2,3,...).
>> >The word 'power' means something because there exist sets S such that the
>> >elements of N can be put in a one-to-one correspondence with a proper
>> >subset of S but the elements of an S cannot be put in a one-to-one
>> >correspondence with a proper subset of N. It has been proven that
>> >aleph-1, the power of the set of all subsets of N (by definition) is
>>
>> That is not the definition of aleph_1. Aleph_1 is simply the next larger
>> cardinal number after aleph_0. The conjecture that aleph_1 = 2^(aleph_0) is
>> called the continuum hypothesis, and is known to be independent of the axioms
>> of ZFC.
>>
>> >greater than the power of N, because the elements of an aleph-0 set can
>> >map into a proper subset of the elements of an aleph-1 set, but not
>> >vice versa.
>>
>> >And it has been proven that defining each aleph number as
>> >the power of the set of all subsets of a set of the previous aleph number
>> >power yields an infinite series of aleph numbers, each larger than the
>> >previous.
>>
>> No, that yields an infinite series of what are called beth numbers. The
>> statement that there are no alephs other than the beth numbers is called the
>> generalized continuum hypothesis.
>>
>> >It has also been proven that aleph-1 cannot be either proven or disproven
>> >to be the same as the power of the continuum, ie. the power of the set
>> >of real numbers.
>>
>> That statement, at least, is correct. However, it is easy to show that the
>> cardinality of the continuum is 2^(aleph_0).
>>
>> >Thus we have Cantorian set theory and non-Cantorian
>> >set theory as equally valid.
>>
>> There is nothing non-Cantorian about any of this.
>>
>> >Now, what in the hell are you people talking about?
David C. Ullrich
***********************
John Sullivan
main thread that was taken before the branch that I thought was the conclusion.
So I will just reiterate that I conceded that I was wrong to jump to the
conclusion that the book in question meant for me to think that the
cardinality of the set of all subsets of a set of cardinality aleph-0 is
the definition of aleph-1.
The reason for my 'Cantorian' remark was that I once read that Cantor had
tried to prove the continuum hypothesis but couldn't, and
when it was finally decided, in the 60s I believe, that the continuum
hypothesis could logically be neither proven nor disproven, that gave
rise to the possibility of 'Cantorian set theory', wherein the continuum
hypothesis is assumed to be true, and 'non-Cantorian set theory', wherein
the continuum hypothesis is not assumed to be true. The analogy given was
that 'Euclidean geometry' is geometry which assumes the postulate that
parallel lines do not meet, and that 'non-Euclidean geometry' is geometry
which does not assume that postulate.
Being under this false impression I thought that I was stating the final
resolution of the continuum hypothesis, which occurred in the 60s I believe,
when I stated that it had been proven that the set of all subsets of a
set of power aleph-0 could be neither proven nor disproven to be the power
of the set of real numbers. Now of course I realize that my statement
of the continuum hypothesis was utterly false. It's true that I did not
label my statement as the continuum hypothesis, but believing it to be
true, I assumed that it would be so recognized.
For the past forty years I have been under the false impression that the
aleph numbers are defined in terms of sets of all subsets, and whenever I
have read about the continuum hypothesis, nothing until now has jolted me
out of that false impression. This thread has been quite a revelation.
The most embarrassing thing about it, however, is not my false impression.
It is that I failed to realize for so long that it is a simple matter to
prove what I had thought erroneously to be the continuum hypothesis,
namely that the cardinality of the set of all subsets of the positive integers
is indeed the cardinality of the set of real numbers.
JS
ull...@math.okstate.edu (David C. Ullrich) wrote in message news:<3b17dac1....@news.okstate.edu>...
Keith Ramsay
su...@altavista.com (John Sullivan) writes:
|For the past forty years I have been under the false impression that the
|aleph numbers are defined in terms of sets of all subsets, and whenever I
|have read about the continuum hypothesis, nothing until now has jolted me
|out of that false impression. This thread has been quite a revelation.
It appears to be a common misimpression. George Gamov, the famous
physicist, wrote a book called _1,2,3 infinity_ which I thought was
mostly pretty good (although it's been awhile since I've looked at it)
but it makes this mistake too.
Sets having cardinality 2^aleph_0 and 2^2^aleph_0 (also called
beth_1 and beth_2) appear more often in mathematics than sets of
cardinality aleph_1 or aleph_2, so in some ways that's the more
familiar sequence of cardinalities.
Keith Ramsay
Keith Ramsay
writes:
|I read in the Mac Tutor history site (St.Andrews, Scotland) that
|L.E.J. Brouwer (early 20-th century) is considered to 'start' topology
|as a discipline (with some famous fixed-point theorem ?-). Also, that
|he later disavowed his own contributions to topology, because he changed
|his mind, by not accepting the 'excluded middle' (only true/false) in
|logic & math-proofs.
What actually happened is interesting. Brouwer had already arrived at
his special philosophy before he did his topological work. He knew
that he would have to do some more "conventional" work, however, in
order to get the kind of position he needed from which to put forth
his own views. He had a teacher who encouraged him in this. I believe
he already knew that his classical topology work wasn't the kind of
intuitionist mathematics he thought we should be doing when he did it.
The Brouwer fixed point theorem says that a continuous mapping from
the unit disk {(x,y):x^2+y^2<=1} to itself always has a fixed point.
The simplest constructively valid alternative is that for any such
map, there exist points in the set which are moved by an arbitrarily
small amount. You might find that the mapping moves a portion of the
disk by a very small amount without necessarily being able to find a
point that isn't moved at all.
Keith Ramsay
Robin Chapman
>Fred Galvin wrote:
>> more or less natural way. The words, of course, are just formal
>> expressions; I don't think there is any sensible way to regard them as
>> infinite *products* of permutations. For example, what permutation is
>> the infinite product d = ccc...? And what is cd? Anyway, the standard
>Thanks Fred, now we're getting somewhere;-) I understand that one
I doubt it. You fail to address Fred's point. How do you interpret
infinte words are permutations?
>with the infinite group Z! = A* with two generators A = {add+1, swap2}.
These generate a proper, countable, subgroup of perm(Z).
>Of course the operation (string concatenation) here means 'permutation
>composition' (of permutations defined formally over Z) which must be
So how do you intend to compose an infinite sequence of permutations?
>You restore my faith in the possibility of having a constructive
>discussion with a mathematician (requiring learning at both sides;-) ,
>which for a while seemed impossible (re: the likes of RC & DU ;-(
If you opened your mind, you could learn a lot from me, but you're
intent on keeping it tightly shut. All I have ever learned from
you is the depth of stupidity of the human race.
ma...@primrose.csv.warwick.ac.uk
Here is an example of a theorem in group theory that involves aleph_1.
Let p and q be distinct primes and let G be an elementary abelian
p-group (i.e. abelian p-group of exponent p). Does there there exist
an elementary abelian q-group N and a nonsplit extension E of N by G?
(That is, a group E having a normal subgroup N with E/N isomorphic
to G, but with no subgroup C such that CN=E and C ^ N = 1.)
The answer (surprisingly) is yes if and only if |G| = aleph_1.
More generally, there exists a G-module N over the field of order q
such that the cohomology group H^n(G,N) != 0 iff |G| = aleph_{n-1}.
Derek Holt.
David C. Ullrich
wrote:
>I thought we had concluded this discussion.
That's usenet. If you look at the dates you see you're
replying to something I wrote two days ago, back before
anything had been "concluded". The post didn't appear for
a while because of the way the internet works.
> Maybe this is some branch of the
>main thread that was taken before the branch that I thought was the conclusion.
>So I will just reiterate that I conceded that I was wrong to jump to the
>conclusion that the book in question meant for me to think that the
>cardinality of the set of all subsets of a set of cardinality aleph-0 is
>the definition of aleph-1.
Yes you conceded that, and everyone was impressed. Seriously - when
someone comes along and starts saying a lot of stuff that's just not
so people wonder whether he's (i) simple mistaken but susceptible to
rational argument or (ii) convinced that he's right and everyone's
wrong no matter _what_ anyone says.
So when you say what you said people wonder - then when you show
that you satisfy (i) that's a pleasant surprise.
>The reason for my 'Cantorian' remark was that I once read that Cantor had
>tried to prove the continuum hypothesis but couldn't, and
>when it was finally decided, in the 60s I believe, that the continuum
>hypothesis could logically be neither proven nor disproven, that gave
>rise to the possibility of 'Cantorian set theory', wherein the continuum
>hypothesis is assumed to be true, and 'non-Cantorian set theory', wherein
>the continuum hypothesis is not assumed to be true.
Fine. I don't think that most people would call this a distinction
between "Cantorian" and "non-Cantorian". And it did take a few
requests from more than one person to get you to explain what
you meant by "non-Cantorian" here (adding to people's concern
over the (i) versus (ii) thing). But if you're simply using
the term in what I _believe_ is a non-standard way, but you're
willing to say exactly what you mean that's no problem.
(See, here on sci.math the people who talk about "non-Cantorian"
this and that are usually people who are insisting that that
theorem in that book of yours is _wrong_. (The theoream about
the power set of S having larger cardinality than S.) So when
you start talking about "non-Cantorian set theory" you _really_
give a funny impression.)
David C. Ullrich
David C. Ullrich
<rjch...@mailandnews.co.uk> wrote:
>>===== Original Message From Nico Benschop <n.ben...@chello.nl> =====
>>Fred Galvin wrote:
>>> more or less natural way. The words, of course, are just formal
>>> expressions; I don't think there is any sensible way to regard them as
>>> infinite *products* of permutations. For example, what permutation is
>>> the infinite product d = ccc...? And what is cd? Anyway, the standard
>
>>Thanks Fred, now we're getting somewhere;-) I understand that one
>
>I doubt it. You fail to address Fred's point. How do you interpret
>infinte words are permutations?
>
>>with the infinite group Z! = A* with two generators A = {add+1, swap2}.
>
>These generate a proper, countable, subgroup of perm(Z).
>
>>Of course the operation (string concatenation) here means 'permutation
>>composition' (of permutations defined formally over Z) which must be
>
>So how do you intend to compose an infinite sequence of permutations?
Someone should point out that while of course this is undefined
in general, it's easy to give examples where there's only one
thing the composition could possibly mean, but the composition
turns out not to be a permutation. (Came up in a thread some time
ago - in that context it was somehow said to be yet one more
proof that Cantor was Wrong.)
>>You restore my faith in the possibility of having a constructive
>>discussion with a mathematician (requiring learning at both sides;-) ,
>>which for a while seemed impossible (re: the likes of RC & DU ;-(
>
>If you opened your mind, you could learn a lot from me, but you're
>intent on keeping it tightly shut. All I have ever learned from
>you is the depth of stupidity of the human race.
So sure enough, you've got a lot more out of it than he has,
justifying his complaint...<g>
>------------------------------------------------------------
>Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
>"His mind has been corrupted by colours, sounds and shapes."
> The League of Gentlemen
>
David C. Ullrich
***********************
Mike Oliver
> Let p and q be distinct primes and let G be an elementary abelian
> p-group (i.e. abelian p-group of exponent p). Does there there exist
> an elementary abelian q-group N and a nonsplit extension E of N by G?
> (That is, a group E having a normal subgroup N with E/N isomorphic
> to G, but with no subgroup C such that CN=E and C ^ N = 1.)
>
> The answer (surprisingly) is yes if and only if |G| = aleph_1.
That is surprising. Could you indicate anything about the
proof?
The particularly strange thing about the result is that if you
force to collapse Aleph_1 (make it countable), you would expect
the nonsplit extension E to remain. I guess it must become
a "split extension" if that's the phrase; i.e. you would add
the subgroup C. But it's sort of surprising that such a
subgroup could be added.
Mike Oliver
>
> On 3 Jun 2001 17:20:57 -0700, su...@altavista.com (John Sullivan)
> wrote:
>> The reason for my 'Cantorian' remark was that I once read that Cantor had
>> tried to prove the continuum hypothesis but couldn't, and
>> when it was finally decided, in the 60s I believe, that the continuum
>> hypothesis could logically be neither proven nor disproven, that gave
>> rise to the possibility of 'Cantorian set theory', wherein the continuum
>> hypothesis is assumed to be true, and 'non-Cantorian set theory', wherein
>> the continuum hypothesis is not assumed to be true.
>
> Fine. I don't think that most people would call this a distinction
> between "Cantorian" and "non-Cantorian". And it did take a few
> requests from more than one person to get you to explain what
> you meant by "non-Cantorian" here (adding to people's concern
> over the (i) versus (ii) thing). But if you're simply using
> the term in what I _believe_ is a non-standard way, but you're
> willing to say exactly what you mean that's no problem.
It's not an unprecedented way of using the term "non-Cantorian".
I think the idea is that it's by analogy with non-Euclidean; those
using the term suggest studying the consequences of (not CH) added
as an axiom.
The trouble is that for anyone but a strict formalist, (not CH) is
just not a very interesting axiom. It has little independent motivation.
That's not the same as no motivation; there are (reasonably) good
intuitive narratives which can be taken as arguing that CH should
fail in the standard (von Neumann) picture of the set-theoretic
universe. (There are also some arguing CH should hold.)
These arguments, though, have in general rather broader
consequences than (not CH). Moreover, in disanalogy to the
case of Euclidean-v-non-Euclidean geometry, they are arguments
about whether a proposition holds in a single intuitively-described
structure (V), not a case of deducing the structure from the
axioms it obeys as with geometry.
D. Ross
| The trouble is that for anyone but a strict formalist, (not CH) is
| just not a very interesting axiom.
I don't think this is entirely true. For example, there was a lot of
interesting activity a while back among general topologists which took
place in the framework of (not CH)+ Martin's Axiom. What I can't recall is
whether (not CH) was ever actually used as such in the proofs, or whether
the point was simply to obtain the results using MA but not using CH.
- David R.
Nico Benschop
>
> On Sat, 2 Jun 2001, Nico Benschop wrote:
>
> > 1. How does the cardinality of Boolean Lattice 2^Z (Cantor's powerset
> > of Z) compare with that of Z! (the group of all permutations of Z)
>
> Yes, card(2^Z) <= card(Z!), and also card(Z!) <= card(2^Z), whence
> card(Z!) = card(2^Z). For the second inequality, a permutation of Z
> can be identified with its graph, which is a subset of ZxZ; and there
> is a bijection between ZxZ and Z.
>
> > 2. Assuming (correct me if I'm wrong) all binary strings, thus over
> > all reals on interval [0,1) (which is an uncountable set;-) ,
> > how come that all strings over {a,b,c} of countable length, where:
> > a= +1, b= -1, c= swap(0,1) leaving rest of Z fixed,
> > (under permutation composition; I take -1, complement of +1,
> > not for granted, although maybe I should, to have just 2 gen's).
>
> Indeed, the set of all infinite words (strings of length omega) over a
> 3-letter (or a 2-letter) alphabet has the same cardinality as Z!; and
> so it is possible to represent elements of Z! as infinite words over
> {a,b,c}; I wouldn't be surprised if it is possible to do so in some
> more or less natural way. The words, of course, are just formal
> expressions; I don't think there is any sensible way to regard them as
> infinite *products* of permutations. For example, what permutation is
> the infinite product d = ccc...? And what is cd? Anyway, the standard
> meaning of the terminology "the group G is generated by its elements
> is equal to some *finite* word in the generators a, b, c. [...]
Re[*]:
Well, we may have to depart here indeed from the standard meaning,
and relax "each element" to "every subclosure" (subgroup). Then
d = ccc... converges to the 2-element cyclic subgroup d = {c,cc} =
{c,e}
where cc=e (the identity element of goup Z! - the "allfix" element)
sothat ccc= c = {all odd-length strings in c*}
and cc = cccc = e = {all even-length ,, }
The claim that {a,b,c}* covers all of Z! would then hold not
element-wise
but subgroup-wise. Of course, each _finite_ string would still yield one
single permutation all right.
BTW: is that so very different from Cauchy's approach towards the reals,
where a _sequence_ of reals defines a limit???
Just replace (real-) "number" by (sub-) "closure", and the rest may hold
still verbatim (overstating it a bit;-).
The concept of "convergence" towards a single element, would have to
relax to: towards (sub-)closure, as I already used in my 'regula falsi'
approach with Cantor's diagonal as a generating device, and
corresponding supra-exponential 2#(r+1) = 2^(2#r), as repeated powerset
order, see:
http://home.iae.nl/users/benschop/cantor.htm
with corresponding transcendental CDT (Cantor-Diagonal Transcendental)
-- NB -- http://home.iae.nl/users/benschop
Mike Oliver
>
> Mike Oliver <oli...@math.ucla.edu> wrote:
>
> | The trouble is that for anyone but a strict formalist, (not CH) is
> | just not a very interesting axiom.
>
> I don't think this is entirely true. For example, there was a lot of
> interesting activity a while back among general topologists which took
> place in the framework of (not CH)+ Martin's Axiom.
I may not have expressed myself clearly. I meant that (not CH) is
not very interesting as an axiom in and of itself, not that it can't
be part of propositions which could be interesting from an axiomatic
approach.
> What I can't recall is whether (not CH) was ever actually used as
> such in the proofs, or whether the point was simply to obtain the
> results using MA but not using CH.
MA follows trivially from CH; if CH is true, then MA gives you nothing.
Specifically, MA says that every infinite cardinal less than 2^Aleph_0
has a certain property which is a generalization of an observation
about Aleph_0. If CH holds, then the only infinite cardinal less than
2^Aleph_0 is Aleph_0 itself, so MA accomplishes nothing.
I think that most of the results in set-theoretic topology along these
lines are proved from MA(Aleph_1), which says that Aleph_1 has the property
alluded to above. MA+(not CH) implies MA(Aleph_1). MA(Aleph_1) implies
(not CH), and while it doesn't imply full MA, it is probably enough to
get most of the results you're talking about.
Bill Taylor
|> What actually happened is interesting. Brouwer had already arrived at
|> his special philosophy before he did his topological work. He knew
|> that he would have to do some more "conventional" work, however, in
|> order to get the kind of position he needed from which to put forth
|> his own views.
Yes, I'd heard about this story before. I wonder if it would still
be the case today, that if someone wanted to get a lot of controversial
but technically sound stuff published, he'd have to "make a rep" first?
I rather think not - we are more amenable to off-stream stuff these days.
The Brouwer situatuion is rather akin to what happens in the abstract art
world. If someone wants to make a real name for himself painting (what most
of us would regard as) talentless rubbish, he first has to make a fair name
painting (at least technically) competent, even expert stuff. Only when he
has shown beyond doubt he "knows where he's putting his brushes" can his
rubbish be taken as "profound", rather than just sloppy daubings.
Thank bog we work in a field where there is at least some objectivity
regarding technical worth.
-------------------------------------------------------------------------------
Bill Taylor W.Ta...@math.canterbury.ac.nz
-------------------------------------------------------------------------------
In science - one tries to tell people, so as to be understood by everyone,
something no one ever knew before. In literature, it's the exact opposite.
-------------------------------------------------------------------------------
David C. Ullrich
wrote:
>"David C. Ullrich" wrote:
>>
>> On 3 Jun 2001 17:20:57 -0700, su...@altavista.com (John Sullivan)
>> wrote:
>>> The reason for my 'Cantorian' remark was that I once read that Cantor had
>>> tried to prove the continuum hypothesis but couldn't, and
>>> when it was finally decided, in the 60s I believe, that the continuum
>>> hypothesis could logically be neither proven nor disproven, that gave
>>> rise to the possibility of 'Cantorian set theory', wherein the continuum
>>> hypothesis is assumed to be true, and 'non-Cantorian set theory', wherein
>>> the continuum hypothesis is not assumed to be true.
>>
>> Fine. I don't think that most people would call this a distinction
>> between "Cantorian" and "non-Cantorian". And it did take a few
>> requests from more than one person to get you to explain what
>> you meant by "non-Cantorian" here (adding to people's concern
>> over the (i) versus (ii) thing). But if you're simply using
>> the term in what I _believe_ is a non-standard way, but you're
>> willing to say exactly what you mean that's no problem.
>
>It's not an unprecedented way of using the term "non-Cantorian".
Oh. Thanks.
>I think the idea is that it's by analogy with non-Euclidean; those
>using the term suggest studying the consequences of (not CH) added
>as an axiom.
>
>The trouble is that for anyone but a strict formalist, (not CH) is
>just not a very interesting axiom. It has little independent motivation.
I may be a fairly formal sort of guy, but I think I'd find
CH interesting in any case. The independence of CH is certainly
interesting because there are various counterexamples in analysis
that depend on CH - these examples do not prove (to me<g>) that the
theorem in question is false, but the examples does show thatit's a
waste of time to try to prove the theorem, which can be a useful
thing to know.
>That's not the same as no motivation; there are (reasonably) good
>intuitive narratives which can be taken as arguing that CH should
>fail in the standard (von Neumann) picture of the set-theoretic
>universe. (There are also some arguing CH should hold.)
>
>These arguments, though, have in general rather broader
>consequences than (not CH). Moreover, in disanalogy to the
>case of Euclidean-v-non-Euclidean geometry, they are arguments
>about whether a proposition holds in a single intuitively-described
>structure (V), not a case of deducing the structure from the
>axioms it obeys as with geometry.
David C. Ullrich
Dave L. Renfro
[sci.math Mon, 4 Jun 2001 09:32:46 +0000 (UTC)]
<http://forum.swarthmore.edu/epigone/sci.math/stantaxfan>
wrote, in part and quoting Keith Ramsay (04 Jun 2001 04:04:42 GMT):
>> Sets having cardinality 2^aleph_0 and 2^2^aleph_0 (also called
>> beth_1 and beth_2) appear more often in mathematics than sets of
>> cardinality aleph_1 or aleph_2, so in some ways that's the more
>> familiar sequence of cardinalities.
>
> Here is an example of a theorem in group theory that involves
> aleph_1.
There are well-known examples in descriptive set theory. Of course,
the Borel hierarchy for the Borel sets has (ordinal) length aleph_1.
But there's a better example that grew out of attempts to prove CH.
In trying to prove CH, early researchers focused on nicely definable
classes of sets when a proof for arbitrary sets wasn't forthcoming.
Recall the Borel hierarchy --->>>
closed ---> F_sigma ---> F_sigma-delta ---> ...
open ---> G_delta ---> G_delta-sigma ---> ...
We say that a class of sets satisfies CH (or the sets individually
satisfy CH) if each of the sets has countable cardinality or
cardinality c.
A stronger property, in the sense that for some models of ZFC a
class of sets can satisfy CH but not satisfy what I'm about to
describe, is that every uncountable set in the collection contains
a nonempty perfect subset. The fact that this is actually a stronger
property was proved by Godel. Godel showed in 1938 that every subset
of the reals can satisfy CH and yet this perfect set property can
fail for some uncountable projective sets (see below). [More
precisely, Godel showed that the axiom of constructibility implies
there exists an uncountable co-analytic set with no nonempty
perfect subset.] Many of the results I mention below for CH were
actually proved by verifying this stronger perfect set property.
One could argue that Cantor showed open sets satisfy CH in 1873-74
when he introduced cardinality, since it is easy to see that every
open interval has the same cardinality as the reals. Around 1882
Cantor proved that closed sets satisfy CH. It immediately follows
that the F_sigma sets also satisfy CH. In 1903 W. H. Young proved
that the G_delta sets satisfy CH. From this it immediately follows
that the G_delta-sigma sets also satisfy CH.
Nothing further was known until Hausdorff and Aleksandrov
(independently) proved that every Borel set satisfies CH,
around 1915-16.
In 1916 Suslin found an error in Lebesgue's influential 1905 paper
on Borel sets (and on various definability issues). Lebesgue
claimed that the projection of any Borel set is a Borel set while
proving something else. [The statement Lebesgue gave an incorrect
proof of wound up being correct, though. This was shown by Lusin in
1917.] Suslin introduced a wider class of sets obtained by looking
at all continuous images of Borel sets. [It is equivalent to just
consider projections from R^2 into R of G_delta sets in R^2.] I think
Suslin originally called them A-sets (for Aleksandrov, I believe),
but after a while they became known as analytic sets.
Suslin proved in 1916 that the analytic sets satisfy CH.
In 1925 (by this time, at least) Lusin introduced the hierarchy
of projective sets. Let 'A', 'C', and 'P' represent analytic,
complements of, and projections of (equivalently in the present
context, continuous images of). Then the projective hierarchy is
A ---> PCA ---> PCPCA ---> ...
CA ---> CPCA ---> CPCPCA ---> ...
Note: This can be continued through the countable ordinals by
taking unions at the limit ordinals. Even if you do this
up to aleph_1 (the ordinal), you're still quite a bit away
from considering ALL subsets of the reals, since you're still
only looking at c subsets of the reals. [The proof is just
like the proof that there are c Borel sets.]
The CA sets are also called the co-analytic sets.
Souslin (1916, I think) showed that the CA sets are either countable,
have cardinality aleph_1, or have cardinality c. [What he actually
proved is that every CA set is a union of aleph_1 many Borel sets,
from which the CH property immediately follows.]
Around 1925-26, Lusin and Sierpinski proved that PCA sets can have
cardinality < aleph_1, = aleph_1, or = c. [For a proof, see around
pp. 518-520 of Jech's "Set Theory", 1978.]
I know that it is consistent in ZFC for CPCA sets to have a variety
of cardinalities, but I don't know which ones are possible. Probably
every aleph_n for finite n's can arise, but I don't know for sure
if even this is the case.
CH can be extended to the higher projective classes with certain
(not-obviously-connected-to-CH axioms). For example, Solovay proved
in 1969 that if there exists a measurable cardinal, then the PCA
sets satisfy CH. [In fact, he showed they have the stronger perfect
set property I mentioned earlier.] Martin showed sometime in the
1970's that the existence of a measurable cardinal implies that every
PCPCA set has cardinality < aleph_1, = aleph_1, = aleph_2, or c.
Woodin showed in 1984 that if there exists a supercompact cardinal,
then every projective set has CH. [The perfect subset property, and
some other nice properties, in fact.]
Dave L. Renfro
D. Ross
| > What I can't recall is whether (not CH) was ever actually used as
| > such in the proofs, or whether the point was simply to obtain the
| > results using MA but not using CH.
| I think that most of the results in set-theoretic topology along these
| lines are proved from MA(Aleph_1), which says that Aleph_1 has the property
| alluded to above. MA+(not CH) implies MA(Aleph_1). MA(Aleph_1) implies
| (not CH), and while it doesn't imply full MA, it is probably enough to
| get most of the results you're talking about.
Thanks for the reminder. I'm sure I knew this at one time; my memory in the
area is suffering from my not having discussed MA the last time I taught our
graduate class in set theory.
BTW, on the same thread (but not at really related to MA), anyone interested
in nifty consequences of CH could do worse than read Sierpinski's very
entertaining little book on CH.
- David R.
Mike Oliver
>
> On Mon, 04 Jun 2001 14:59:04 -0700, Mike Oliver <oli...@math.ucla.edu>
> wrote:
>> The trouble is that for anyone but a strict formalist, (not CH) is
>> just not a very interesting axiom. It has little independent
>> motivation.
>
> I may be a fairly formal sort of guy, but I think I'd find
> CH interesting in any case.
It looks like I didn't express myself very well as no one seems
to have understood me. I certainly didn't mean *CH* was uninteresting!
I consider it the premier unsolved problem of set theory. BTW
there was a four-day conference at Berkeley last week on CH; I
was up there but kind of sick and only managed to make it to half
of one talk.
I mean that CH and (not CH) are not very interesting as
*candidate*axioms*. *In*and*of*themselves* they don't say
very much, don't correspond one-for-one to insights about
the set-theoretic universe. There are such insights that
might *imply* CH or (not CH), but they imply other things as well.
As a rough analogy, suppose that FLT is not provable from Peano
Arithmetic (still an open question AFAIK). Would you think of
FLT as a candiate axiom for arithmetic? Oh, you might be interested
in the consequences of PA+FLT, but I think that line of inquiry
is rather limited. The *reason* you believe in FLT is because
it follows from arguments involving structures not formalizable
in PA, and the interesting thing to do is to incorporate those
structures into your outlook, not simply to adopt FLT as a formal
axiom.
David C. Ullrich
wrote:
>"David C. Ullrich" wrote:
>>
>> On Mon, 04 Jun 2001 14:59:04 -0700, Mike Oliver <oli...@math.ucla.edu>
>> wrote:
>>> The trouble is that for anyone but a strict formalist, (not CH) is
>>> just not a very interesting axiom. It has little independent
>>> motivation.
>>
>> I may be a fairly formal sort of guy, but I think I'd find
>> CH interesting in any case.
>
>It looks like I didn't express myself very well as no one seems
>to have understood me. I certainly didn't mean *CH* was uninteresting!
>I consider it the premier unsolved problem of set theory. BTW
>there was a four-day conference at Berkeley last week on CH; I
>was up there but kind of sick and only managed to make it to half
>of one talk.
I've got pretty sick every time I've attended a conference on CH.
>I mean that CH and (not CH) are not very interesting as
>*candidate*axioms*. *In*and*of*themselves* they don't say
>very much, don't correspond one-for-one to insights about
>the set-theoretic universe. There are such insights that
>might *imply* CH or (not CH), but they imply other things as well.
>
>As a rough analogy, suppose that FLT is not provable from Peano
>Arithmetic (still an open question AFAIK). Would you think of
>FLT as a candiate axiom for arithmetic? Oh, you might be interested
>in the consequences of PA+FLT, but I think that line of inquiry
>is rather limited. The *reason* you believe in FLT is because
>it follows from arguments involving structures not formalizable
>in PA, and the interesting thing to do is to incorporate those
>structures into your outlook, not simply to adopt FLT as a formal
>axiom.
Ok. (You must be aware the can of worms you're opening here,
I've-found-a-simple-proof-of-FLT-wise. Probably should
choose your analogies more carefully<g>.)
David C. Ullrich
***********************
Bill Taylor
> Specifically, MA says that every infinite cardinal less than 2^Aleph_0 has
> a certain property which is a generalization of an observation about Aleph_0.
Mike - please tell us - what *is* that property?
I've often idly wondered what Martin's axiom was, this looks like a good time
to find out.
And what is the motivation for it? The way you describe it, it seems to be
just a cunning way to get some nice consequences of CH without having to
admit to the odium of actually believing it.
Is that about the size of it??
-------------------------------------------------------------------------------
Bill Taylor W.Ta...@math.canterbury.ac.nz
-------------------------------------------------------------------------------
In math we decide on the rules, then try to deduce the consequences.
In science we observe the consequences, then try to deduce the rules.
-------------------------------------------------------------------------------
Mike Oliver
>
> Mike Oliver <oli...@math.ucla.edu> writes:
>
> > Specifically, MA says that every infinite cardinal less than 2^Aleph_0 has
> > a certain property which is a generalization of an observation about Aleph_0.
>
> Mike - please tell us - what *is* that property?
MA(kappa) means that for any ccc partial order and any kappa-many dense
open subsets from the p.o., there's a filter that meets all those open
subsets.
> I've often idly wondered what Martin's axiom was, this looks like a good time
> to find out.
>
> And what is the motivation for it? The way you describe it, it seems to be
> just a cunning way to get some nice consequences of CH without having to
> admit to the odium of actually believing it.
Martin's original motivation was the notion that cardinals less than the
size of the continuum should somehow be very similar to Aleph_0. He
later recanted that intuition, said he didn't know why he ever thought
that. But in any case, one way MA is commonly used is something
the formalists would kind of like, I suppose: For many questions
in set-theoretic topology, you can show that ZFC+MA(Aleph_1) proves
one answer to the question, and some combinatorial principle like Diamond
proves the other answer. Once you have this situation you know that
the question is independent of ZFC, because both MA(Aleph_1) and Diamond
are relatively consistent with ZFC.
FGD
>
> Mike Oliver <oli...@math.ucla.edu> writes:
>
> > Specifically, MA says that every infinite cardinal less than 2^Aleph_0 has
> > a certain property which is a generalization of an observation about Aleph_0.
>
> Mike - please tell us - what *is* that property?
I'm not Mike, but I'll answer anyway...
> I've often idly wondered what Martin's axiom was, this looks like a good time
> to find out.
>
> And what is the motivation for it? The way you describe it, it seems to be
> just a cunning way to get some nice consequences of CH without having to
> admit to the odium of actually believing it.
There are many ways to look at MA, I'll take a more combinatorial
approach. When trying to construct objects with certain properties, we
often resort to an inductive construction by successive approximations.
So in order to construct an object T with properties P_0, P_1,..., we
start with a first approximation T_0 which has property P_0, then we
extend T_0 to a better approximation which satisfies P_0 and P_1, and so
on. Finally we take T to be the limit of T_0, T_1,... which will then
have all properties P_0, P_1,...
We can axiomatize this construction as follows. We first organize the
approximations in a set A which we order by extension so that A = (A,<)
is a partial order with T < T' denoting "T' extends T". (Note 1) We
assume that < is reflexive and transitive. So our above sequence T_0,
T_1,... is just an increasing chain in A. The limit T is not required to
be in A since in the general case the approximations and the disired
objects will be wildly different, e.g. the approximations can carry with
them a lot of auxiliary information which is not necessary in the final
object.
Having so organized our approximations, we may look at the properties
P_0, P_1,... as subsets of A. But in order for the above construction to
work, the properties have to be nice enough. For one thing, we want the
properties to be preserved under extension so that once we get an an
appoximation with property P, all future approximations also have this
property. We also want to be able to reach all of the properties so that
from any point T in A we can find an extension T' of T with property P.
To summarize, we want our sets P to satisfy:
(1) If T \in P and T' > T, then T' \in P.
(2) For all T, there is a T' \in P with T' > T.
A subset P of A which satisfies (1) is called open, and one which
satisfies (2) is called dense. So we want all of our sets P_n to be open
dense.
This is enough to make the above construction work:
THEOREM MA_omega. If A = (A,<) is a partial order and D_0, D_1,... are
open dense subsets of A, then theste is an increasing chain T_0 < T_1 <
... in A which eventually belongs to each D_n.
This is the basic form of a typical Martin's Axiom (Note 2), the
subscript omega denotes the number of open dense sets in the statement.
Already, this basic theorem has some interesting consequences, for
example
COROLLARY. The real numbers are uncountable.
Proof. If we look at real numbers as binary sequences. The natural
approxmation is a finite binary sequence ordered by end extension. Let A
= (A,<) be this partial order. We show that given any reals r_0, r_1,...
we can find a real r distinct from all of them. To do this we define the
following subsets of A: let D_n be the set of all finite binary
sequences t of length at least n which differ from r_n at some digit. If
t belongs to D_n, then any real r extending t must be distinct from r_n
for if t and r_n differ in the i-th digit, then so will r. The length
condition is to guarantee that the limit of our chain will be infinite.
So if the chain t_0 < t_1 < ... eventually belongs to each D_n, the
limit r will be a real different from all r_n.
It thus suffices to show that the D_n are all open dense. The fact that
they are open is clear. For dense, if t is a binary sequence of length
m, then extend it to a sequence t' of lenth m+1 whose last digit is
different from the corresponding digit of r_n, and then extend t' by
tagging on sufficiently many 0's so that the resulting sequence t* has
length at least n. Then t* is an extension of t in D_n as required. #
The above proof generalizes to prove that 2^k > k for every
infinite cardinal k. In that case, we use the partial order A_k of all
binary sequences of length less than k ordered by end extension. Note
that if k is regular, any chain of length less than k has an upper bound
in A_k. A partial order with that property is said to be k-closed. We
can
easily prove another Martin's Axiom.
THEOREM MA_k(k-closed). If A = (A,<) is a k-closed partial order and
D_i, i < k, are open dense subsets of A, then there is an increasing
chain {t_i: i < k} in A which eventually belongs to each D_i.
Note that every partial order is omega-closed (every finite chain has an
upper bound), so MA_omega is the special case k=omega of the above. With
this, the proof of the above corollary goes through to show that 2^k > k
for each regular k. For singular k, a similar argument can be used to
prove Koenig's theorem k^cf(k) > k which then implies 2^k > k. The above
Martin's Axiom are theorems of ZFC yet they are already very useful,
though one generally prefers a direct inductive construction. Can we
improve these? Not in general, our previous arguments to prove 2^k > k
easily yield
FACT. MA_m(X) implies 2^k > m where X is any class of partial order
containing the partial order A_k of all binary sequences of length less
than k ordered by end-extension.
So under GCH, MA_m(k-closed) is false for all m > k which shows that k
is optimal for plain ZFC without further assumptions. In fact,
MA_m(k-closed), m > k, is plain false unless further restrictions are
imposed on the partial orders. (Note 3) So the best we can hope for is
something MA_m(k-closed+???) and the best we can get is a relative
consistency result if A_k satisfies ???.
In the above MA, we have used chains to meet the target open dense sets.
This is not suitable for all partial orders. For the general case we
must use filters: a subset F of the partial order A is a filter if
(1) If T \in F and T' < T, then T' \in F.
(2) If T, T' \in F, then there is a T* \in F with T, T' < T*.
Two elements T, T' of A which have a common extension are said to be
compatible, they are incompatible when they have no common extension.
Property (1) is not essential, for if F satisfies (2), then its downward
cloure satisfies (1) and (2). A chain obviously satisfies (2) for we can
take T* to be the greatest of T and T'. So every chain leads to a
filter, but the converse is not true. With filters, the generic MA
becomes MA_k(???): If A = (A,<) is a partial order with property ??? and
D_i, i < k, are open dense subsets of A, then there is a filter F in A
which meets each D_i.
The original Martin's Axiom generalizes MA_omega, the additional
restriction on the partial orders is the countable chain condition,
or ccc for short. (Note 4) A subset B of the partial order A is said to
be an antichain if any two elements of B are incompatible. We say that A
satisfies the ccc if every antichain of A is countable, equivalently
every uncountable subset of A contains two compatible elements. The
partial order A = A_omega which we used to prove that the reals are
uncountable satisfies the ccc. Indeed, A is countable so any subset of
A, antichain or not, is also countable. So it follows from the Fact that
MA_m(ccc) is false when m >= 2^omega. So the best we can hope for is
MA_m(ccc) for all m < 2^omega.
DEFINITION. The original Martin's Axiom, denoted MA(ccc) or simply MA,
is the statement that MA_m(ccc) holds for all m < 2^omega.
Of course, if CH holds, then MA is trivially true, so it is customary to
use Martin's Axiom only for MA(ccc) + not CH. It has been shown that if
ZFC is consistent, then so is ZFC + MA(ccc) + not CH. MA has many
applications, the most famous is without doubt the fact that it implies
Suslin's Hypothesis. But there are many lesser, but perhaps more useful
results in many branches of mathematics.
What about generalizations of MA_k(k-closed) for uncountable k? We would
hope that MA(k-closed + k-cc) is consistent where k-cc, the k-chain
condition, is the direct generalization of the ccc, namely: every
antichain of has size at most k. (Note 5) However, the problem is much
more difficult than that and no satisfactory answer is known for large
k. For k = omega_1, we have some interesting results. Baumgartner has
given a simple proof that MA_{omega_2}(omega_1-compact + omega_1-linked)
is consistent where omega_1-compact and omega_1-linked are strong forms
of omega_1-closed and omega_1-cc respectively, and this has been
slightly improved by others.
But the most far reaching generalization of MA is certainly the Proper
Forcing Axiom or PFA which came shortly after Shelah's discovery of
proper forcing. We shall not discuss here what proper partial orders
are, but we mention that PFA is generalizes both MA and
MA_{omega_1}(omega_1-closed) in a significant manner. Following the PFA
breakthrough, there have been substantial developments by Shelah,
Todorcevic, Woodin, and others so that low cardinal forcing axioms are
very well understood today though our understanding is still incomplete.
For higher cardinals, the situation is much less clear and would require
another major breakthrough in forcing.
That's all for now, I hope you enjoyed this very brief introduction to
forcing axioms. Some parts are missing, like the motivation for ccc and
references and history. (I'm traveling so I don't have much with me
besides memory...) The PFA stuff needs expansion including mention such
things as Martin's Maximum, the role of stationary sets and maybe Pmax
and more of Woodin's stuff. I would also like to give a non-trivial
application of MA such as proving that Lebesgue measure is
2^omega-additive, but I didn't have time to write down a proof. Perhaps
some of this will come in a later revision....
-- Frank
----
Note 1: Usually one sees the counter-intuitive t' < t for "t' extends t"
instead of t < t' as is customary in forcing. This comes from Boolean
algebras where it makes sense to look at 1 as the element with least
information since it is contained in every filter.
Note 2: The original Martin's Axiom is what we denote here by MA(ccc).
The other axioms have been called by various names: Generalized Martin's
Axiom, Extended Martin's Axiom, Martin-Like Axioms, etc. We have avoided
these variations to underline the uniformity of this family of
statements.
Note 3: Consider the partial order A_k(X) of all X-ary sequences (like
binary sequences, but whose digits elements of X) with length less than
k where k is regular. Then A_k(X) is k-closed. Show that the sets D_x =
{t: x is a digit of t}, x \in X, are all dense in A_k(X). Deduce that
any chain of A_k(X) which meets all D_x leads in the limit to a
surjective map T:k->>X. So if |X| <= m and MA_m(k-closed) holds, then
|X| <= k. Conclude that MA_m(k-closed) holds if and only if m <= k. Show
that A_omega(X) does not satisfy the ccc if X is uncountable.
Note 4: The countable chain condition should really be called the
countable antichain condition, and some do call it such. This also comes
from complete Boolean algebras where the countable chain condition (with
the obvious meaning) and the countable antichain condition are
equivalent. This is not true for general partial orders, e.g. a chain,
an antichain.
Note 5: (We should really call this the (k^+)-cc instead of the k-cc,
but this is ASCII and the simpler notation is better for my typing.) It
can be shown that any separative k-closed partial order does not have
the m-cc for any m < k. In fact it does not have the m-cc whenever m <=
2^n for some n < k. This is clear in the case of A_k where the sequences
of same length form an antichain.