From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Subject: Re: Axiom of Choice Debate
Date: 20 Feb 1999 20:32:03 -0500
Newsgroups: sci.logic,sci.math
In article <36CE84B7.77@rochester.rr.com>,
Jonathan W. Hoyle wrote:
:I am having difficulty understanding the argument against the use of the
:Axiom of Choice. In particular, let us consider the classic example
:from Analysis I of producing a non-measurable set. In this
:construction, the line segment [0,1] has been partitioned into
:uncountably many disjoint subsets, and the A.C. is invoked to create a
:set containing one element from each set.
:
:By the end of this proof, the set created via A.C. is shown to be
:non-measurable. Those who argue against the use of the Axiom of Choice
:want to prevent me from constructing such a set. YET DOESN'T THIS SET
:ALREADY EXIST WITH OR WITHOUT THE AXIOM OF CHOICE? After all, it is a
:member of the power set of real numbers, P(R), and thus it really
:doesn't matter HOW the set was created. It only matters that this set
:already exists (as a member of P(R)), and must be reckoned with.
:
:Can anyone out there who disagrees with the Axiom of Choice possibly
:explain why to me??
If there is a model for the theory of sets (for that, the theory must be
consistent - and that is impossible to prove) then inside such a model
there is a "minimal" model (Goedel's delta model) which is a Platonist's
paradise: the axiom of choice becomes a theorem, so does the generalized
continuum hypothesis, and every (uncountable) cardinal number is
accessible, and many other comforting circumstances. Moreover, up to
isomorphism, such a minimal model is unique. In delta model, your prayers
are answered.
Then there are models (of ZF) in which AC fails, and one can to certain
extent control how badly it fails.
One way of looking at it: One somehow (and consistently) adjoins a lot of
new elements but not enough functions, so that the choice functions
cannot have arbitrary domains.
Another way: one evaluates the sentence "x belongs to y" not in the "yes
or no" Boolean algebra that we love so much but in another Boolean
algebra, and then the truth value is reduced through an ultrafilter.
The problem of choice reduces to the complete distributivity of that
Boolean algebra, and algebraists know examples of Boolean algebras which
are not completely distributive, If they are careful enough, an
ultrafilter will not erase that failure.
And I think I remember that AC can be violated this way to make all
subsets of R measurable (several authors, I remember Tomas Jech).
Cheers, ZVK(Slavek).
==============================================================================
From: hrubin@odds.stat.purdue.edu (Herman Rubin)
Subject: Re: Axiom of Choice Debate
Date: 20 Feb 1999 08:53:12 -0500
Newsgroups: sci.logic,sci.math
In article <36CE84B7.77@rochester.rr.com>,
Jonathan W. Hoyle wrote:
[as quoted above -- djr]
How about an explanation by someone who neither agrees nor disagrees,
and who has worked on the problem?
One cannot just put down a collection of words to describe a set.
The paradoxes are already there in dealing with sets of integers.
There are only a countable number of expressions in whatever language
we are using to describe a set of integers, so there must exist
sets of integers not described. This is the idea of Cantor's
diagonal argument about the uncountability of the reals.
But even statements about integers may not be meaningful, such as
The smallest integer not defined in one hundred characters.
As this sentence fits in an 80 character line, it leads to a
contradiction, and any contradiction brings down the whole
mathematical edifice.
Both the Axiom of Choice and its negation are consistent.
I believe it is not known whether the non-existence of
non-measurable sets is consistent. It is known that the
real numbers being a countable union of countable sets,
which would give the analysts major problems, is consistent.
If the Axiom of Determinateness holds in a model of set
theory, these analysis problems would disappear, but all
sets would be Lebesgue measurable, and AD is "intuitively"
obvious. AD directly contradicts AC.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
==============================================================================
From: "Robert E. Beaudoin"
Subject: Re: Axiom of Choice Debate
Date: Sun, 21 Feb 1999 00:34:13 -0500
Newsgroups: sci.logic,sci.math
To: Herman Rubin
Herman Rubin wrote:
>
[snip]
> Both the Axiom of Choice and its negation are consistent.
> I believe it is not known whether the non-existence of
> non-measurable sets is consistent. It is known that the
> real numbers being a countable union of countable sets,
> which would give the analysts major problems, is consistent.
> If the Axiom of Determinateness holds in a model of set
> theory, these analysis problems would disappear, but all
> sets would be Lebesgue measurable, and AD is "intuitively"
> obvious. AD directly contradicts AC.
> --
[snip]
Solovay proved (late sixties or early seventies) that
consistency of ZFC + "there exists an inaccessible
cardinal" implies consistency of ZFC + "every set of
reals is Lebesgue measurable." Shelah proved ("Can you
take Solovay's inaccessible away?", _Israel J. Math._, 1984)
that the converse implication is also true. (So "knowing
that the non-existence of non-measurable sets is consistent"
is the same as "knowing that the existence of inaccessible
cardinals is consistent"; to some people it will seem clear
that we already know this, whereas to others such a thing
seems unknowable in principle!)
By the way, why do you call AD "intuitively" obvious? Just
curious.
Robert E. Beaudoin
Sensimetrics Corporation
==============================================================================
From: "Jonathan W. Hoyle"
Subject: Re: Axiom of Choice Debate
Date: Sun, 21 Feb 1999 01:39:58 -0500
Newsgroups: sci.logic,sci.math
I am familiar with the results of ZF + AD yielding that call sets of
real numbers are Lebesque measurable, but I had trouble with that since
the set's non-measurability had nothing to do with AC, only its
construction. What I am beginning to see is that without AC, this set
can't be "specified", and with AD, all such "specifiable" sets are
measurable. Although the Power Set Axiom is available in both cases,
P(R) with AD must be a different power set than P(R) with AC. The
reason all subsets of R are measurable with AD is that the
non-measurable ones aren't "sets" in ZF + AD. Is this correct
reasoning?
If so, it seems that AC generates a more discrete topology for P(R),
whereas AD describes a coarser one. Of course the results of this means
we must reinterpret what "power set" means, since without AC, it is
something very different than what is usually meant.
==============================================================================
From: hbe@sonia.math.ucla.edu (H. Enderton)
Subject: Re: Axiom of Choice Debate
Date: 21 Feb 1999 17:50:42 GMT
Newsgroups: sci.logic,sci.math
Herman Rubin wrote:
>I believe it is not known whether the non-existence of
>non-measurable sets is consistent.
Solovay showed (1970) that this *is* consistent (and you can throw
in DC), assuming Cons(ZFC + exists an inaccessible). That
assumption is not so bad, since the real died-in-the-wool
Platonists figure that ZFC + exists an inaccesible is "true."
(Reference: JSL XXXVIII 529).
--Herb Enderton
hbe@math.ucla.edu
==============================================================================
From: Richard Carr
Subject: Re: Lebesgue Measure vs. Inaccessible Cardinals
Date: Fri, 7 May 1999 23:08:09 -0400
Newsgroups: sci.math,sci.logic
On Fri, 7 May 1999, Jonathan W. Hoyle wrote:
:Date: Fri, 07 May 1999 21:52:31 -0400
:From: Jonathan W. Hoyle
:Newsgroups: sci.math, sci.logic
:Subject: Lebesgue Measure vs. Inaccessible Cardinals
:
:In Ian Stewart's book "From Here to Infinity", the following statement
:is made on p. 69:
:
:"...R. M. Solovay showed that the axiom 'there exists an inaccessible
:cardinal' implies that every set of reals is Lebesgue measurable...and
:subsequently Saharon Shelah proved the converse."
If this is the exact quote then Stewart is incorrect. Solovay's result is
if ZFC+there exists an inaccessible cardinal is consistent then so is
ZF+a whole load of other things :).
One of these things was that every set of reals is Lebesgue measurable.
Although obviously one does not get AC in the model, one does get
countable choice and indeed a bit more.
Details of this can be found, for example, in A. Kanamori (The Higher
Infinite). It was a very early forcing.
Shelah showed that Con(ZF+every set of reals is Lebesgue measurable)
implies Con(ZF(C?)+there exists an inaccessible cardinal).
The paper you'd need to look at is Sh176.
:
:Is this correct? Am I missing something?
As stated it wasn't. You are alert rather than missing something.
:
:Since the Axiom of Choice is all that is required to be able to
:construct a non-Lebesgue measurable set, the above quote implies the
:statement "the Axiom of Choice is true iff there are no inaccessible
:cardinals." Yet in all my readings of inaccessible cardinals, I have
:never heard this before.
Since it is not possible to prove (in ZF(C)) the consistency of the
existence of inaccessible cardinals but it is possible to prove (in ZF)
the consistency of both ZFC and ZF+not AC, then it is no surprise you
haven't heard it. I think that the consistency of an inaccessible cardinal
also implies the consistency of V=L+there is an inaccessible cardinal or
at least of GCH+there is an inaccessible cardinal. GCH implies AC, V=L
implies a very strong version of AC, so that would totally destroy the
supposed equivalence. (The supposed equivalence also imply that it would
not be possible in ZF to prove ZF+not AC consistent, which certainly is
not the case.)
:
:In fact, in "Infinity and the Mind", Rudy Rucker spends 45 pages in
:wonderful detail on the chapter entitled "The Transfinite Cardinals",
:and never once mentions this result, despite describing in detail both
:the Axiom of Choice and inaccessible cardinals.
This is understandable.
:
:Is the term "inaccessible cardinal" being used to describe something
:different in Stewart's book than is traditionally meant?
I'm not sure, having not read Stewart's book. Generally inaccessible is
either going to mean weakly or strongly inaccessible and usually
inaccessible is reserved just for strongly inaccessible (so that if you
want to talk about weakly inaccessible you add the qualifying "weakly".
The nice thing about kappa being strongly inaccessible is that V_kappa
models ZFC (if V does).
(Under GCH, both notions of inaccessibilty are the same, of course.)
Richard Carr.
==============================================================================
From: lpa@netcom.com (Pierre Asselin)
Subject: Re: Lebesgue Measure vs. Inaccessible Cardinals
Date: Sat, 8 May 1999 03:51:05 GMT
Newsgroups: sci.math,sci.logic
"Jonathan W. Hoyle" writes:
>In Ian Stewart's book "From Here to Infinity", the following statement
>is made on p. 69:
>"...R. M. Solovay showed that the axiom 'there exists an inaccessible
>cardinal' implies that every set of reals is Lebesgue measurable...and
>subsequently Saharon Shelah proved the converse."
>Is this correct? Am I missing something?
If the quote is accurate, Stewart goofed. Solovay's result is
that the *consistency* of "there exists an inaccessible cardinal"
implies the *consistency* of "every set of reals is Lebesgue
measurable".
More precisely, let
ZF = Zermelo-Fraenkel
DC = the axiom of dependent choice
IC = there exists an uncountable inaccessible cardinal
LM = every set of reals is Lebesgue measurable
Solovay: Consis(ZF+IC) --> Consis(ZF+DC+LM)
Shelah: Consis(ZF+IC) <-- Consis(ZF+DC+LM)
Reference: Stan Wagon, "The Banach-Tarski Paradox", Cambridge
University Press 1985, ISBN 0-521-45704-1.
--
--Pierre Asselin, Westminster, Colorado
lpa@netcom.com
==============================================================================
From: hrubin@odds.stat.purdue.edu (Herman Rubin)
Subject: Re: Axiom of Choice & Measurability
Date: 1 Nov 1999 20:01:17 -0500
Newsgroups: sci.math
In article <381E0A03.63B2@kodak.com>, Jonathan Hoyle wrote:
>It is well known that denying the Axiom of Choice results in all subsets
>of the reals being measurable. That is, the class of reals which is an
>unmeasurable set within AC, is not a set at all (and thus a proper
>class) with ~AC. My question is: are all measurable subsets of R with
>AC, still sets with ~AC?
There are models without choice which also have non-measurable
sets. It is AD, the Axiom of Determinateness, which guarantees
no non-measurable sets. There are lots of ways of dropping the
Axiom of Choice, and few of them force all subsets of the reals
to be measurable. This is a very strong property.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558