Re: GCH vs. Axiom of Choice.

From: KRamsay (kramsay_at_aol.com)
Date: 12/24/04 

Date: 24 Dec 2004 08:38:25 GMT

In article <1103745661.488813.34230@z14g2000cwz.googlegroups.com>, "J.E."
<troubled6man@yahoo.com> writes:
|The background is that people like Mike Oliver want to say that "all
|subsets are sets", which in a vacuous sense I undestand is true, since
|being a set is a part of being a subset. But either the axiom of
|separation is vacuous or it has content. If it has content, then the
|assumption is that some things are not sets,

I don't see the point of what you're saying here. Of course some
things (e.g. keyboards, chairs) are not sets. But why does that
seem related to the axiom of separation?

|and specifically the axiom
|of separation only guarantees the existence of countablely many
|subsets, since the formulas are countable.

I think you're assuming more here than you're aware of.

Consider a somewhat technical but important feature of the axiom
scheme of separation. The predicate doing the separating is allowed
to have parameters in it. If P(x,y) is a formula with two free
variables in it, one instance of the scheme is

   For every A, for every x, there exists a set whose members
   are the members y of A satisfying P(x,y).

There is not just one set being asserted to exist here. As x varies
the subset of A varies, and the axiom asserts they all exist.

In the diagonal argument, we have such an application of the separation
axiom. For each function f on the integers whose values are sets of
integers, there exists a set {n : n is not a member of f(n)}-- so says
one instantiation of the separation scheme. This implies that for each
function f from N to P(N) there exists an element of P(N) not in the
image, which is what uncountable means.

There are lots of seductive hand-waving arguments related to this, and
if you want to get clear on things, you need to be careful not to get
taken in by them.

| The standard interpretation
|is that the power set contains "more" subsets than the axioms of
|separation required on their own, because "all" is supposed to include
|"more" than countably many.

I don't know whether someone has been saying this, but this is not
how I would describe the situation.

| If people want to say that powers makes
|separation vacuous, that's one thing,

I don't know anyone who would say that either.

| but ignoring that and yet
|insisting that the power set contains all the subsets and that all
|subsets are in the power set

That's the definition of "power set".

|seems to undermine the whole point of
|separation, since the naive assumption that a set is "anything that
|isn't too big" just takes over wholesale.

No, the power set axiom just asserts that there exists a set containing
all (and only) the subsets of a set. The power set axiom is consistent
with pretending that there do not exist (say) uncomputable subsets of the
integers (which is not something anybody I know of claims, by the way).
You need the axiom of replacement or something like that.

|And I think it's unfair to expect you to defend Mike's views that that
|was reasonable, but if you choose to defend it as reasonable, then I
|want an explanation.

I don't know what Mike you're referring to, or which of his views
you have in mind.

| If Pn for n=0,1,... are "all" the one-place
|predicates,

In the language of ZFC, perhaps?

| then is the collection {n in collection iff "it is not the
|case that n satisfies P"} a set? If so, what's the proof?

{n : ~Pn(n)}? This is undefinable in the language of ZFC (if that's
the language whose predicates you're enumerating) because of
Tarski's undefinability of truth. So it certainly can't be proven
to exist in ZFC. Do I think it's a set? Yes. It's possible to make
a (relatively minor) extension of ZFC to give it a predicate for the
truth of the language of ZFC, and allow predicates in the extension
in the selection axiom scheme and so on. I think such an extension
is correct. There's no further justification besides the fact that
it seems intuitively plausible.

I would note that when we say that { { m : Pn(m)} : n is an
integer} is a countable collection of sets-- those selected by
one-place predicates in the language of ZFC-- we are making appeal
to the same kind of extension of ZFC. The argument that the sets
definable in ZFC are countable in number is not expressible in the
language of ZFC, let alone provable from the axioms of ZFC. I
accept both of these facts (the countability of the sets definable
in ZFC and the existence of a diagonal set for the sets of naturals
definable in ZFC) for the same reasons.

| I'm
|assuming that it's some kind of replacement, but repalcement just uses
|two-place predicates, still countable, so can't we just let Pn(x,y) be
|those predicates and make a new diagonal subset. A proof by
|contradiction doesn't show that a different subset was in the power
|set, only that the theory can be so impoverished as to lack the assumed
|set.

I don't really follow what you're saying.

|My only point is that saying "all sets" is vague, since clearly not all
|things are sets.

I don't see what you think your reason for thinking this is. Not
all things are bananas; is there a problem with asking questions
about "all bananas"? It seems like you're thinking in terms of
some picture where we have a bunch of things that might or might
not be allowed as sets, and then we pick out some of them to be
sets. But really I can't tell why you think your claim here makes
sense.

| But this gets back to many vague things. For
|instance, the smallest subset of the set asserted by the axiom of
|infinity to exist, how do we know it doesn't contain anything besides
|the finite ordinals?

It's basically the definition of "finite ordinal".

| Like a nonstandard integer? The nonstandard
|integers exist in all standard sets that contain an infinite subset of
|integers, so they are in the smallest set that contains all the finite
|integers too.

So now you've read some nonstandard analysis, eh?

There are a few ways of defining "nonstandard", and how to explain
whether the integers contain nonstandard integers and how we know
depends on which definition is being used. Sometimes it's a matter
of the author making a collection of assumptions that are
consistent with ZFC but contrary to the way most people interpret
its language. Often, though, it's based on a gnarly interpretation
of the language of ZFC, which we can accept as valid without any
conflict with the original one. But to some degree we need to keep
them separate.

Keith Ramsay

Relevant Pages

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