Re: GCH vs. Axiom of Choice.
From: KRamsay (kramsay_at_aol.com)
Date: 12/26/04
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Date: 26 Dec 2004 18:42:39 GMT
J.E. wrote:
|[I wrote:]
|> [J.E. wrote:]
|> |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.
|
|The axiom of separation says that for every existing sentence and set,
|there is at most one new set.
No, it doesn't. This is fuzzy terminology. What does "new" mean,
after all? It would make more sense if you wrote
The axiom of separation only asserts the existence of one set
for every existing sentence and set.
You wrote of how it "restricts". It doesn't "restrict". It merely fails
to assert more than it actually does.
But actually, the axiom scheme of separation doesn't talk about
sentences. This is a subtle but important point, because if the
axiom scheme actually asserted the existence of a set for each
formula of the appropriate kind, it would mean that we had in it
a defined correspondence between formulas and sets.
It doesn't define such a correspondence. Instead, for each
formula with at least one free variable, we have a separate instance
of the axiom scheme. The axiom scheme instance merely asserts
the existence of one set for each n-tuple of sets.
|> 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.
|
|This is just question begging.
No, it's not.
|If somehow you have more sets than
|sentences, then maybe this "process" could create more subsets than
|predicates, but that's entirely assuming the conclusion. Hardly
|convincing to anyone.
But this is not an argument I made. The idea that there is a
"process" is your idea. I think I know what you're vaguely alluding
to, but you haven't explained it with any precision.
The fact that some instances of the axiom scheme of separation
assert that for each set f there exists a set S bearing a certain
relation to f is simply correct, and although it seems irrelevant
to you, is in fact relevant.
|> 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.
|
|For every function, ... that is also a set.
Note that the premise as well as the conclusion has a universal
quantifier over sets. You didn't object to it before.
| That's where the whole
|arguement becomes viciously circular.
Not at all.
| We both already know that
|separation excludes things from being sets that otherwise could be.
|And we probably both already know that excluding sets can make things
|"uncountable" that otherwise wouldn't be, for instance, let
|V=U{N,{N,U{N,{N}}}} (in binary union notation V=NU{N}U{NU{N}}. Then
|NU{N} is "uncountable", which means more about the sparseness than
|about NU{N} as a collection. Because if V=L, then NU{N} is countable.
The fallacy here is that you are trying to dispute a proof of a
fact A->B by interpreting the premise A and the conclusion B in
different ways. In this case, the premise A is an instance of the
axiom scheme of separation, and the conclusion B is the fact that
there are uncountably many reals. Each of them has a universal
quantifier (over sets) in front.
You are explaining how if the universal quantifier in B is restricted
to only certain selected sets, then it doesn't necessarily mean the same
thing as if the universal quantifier is less restricted. This is true. But
it doesn't invalidate the argument, because if you take the universal
quantifiers in A and B to range over the same class, the implication
holds. It only seems wrong because you first interpret the universal
quantifier in the instance of the axiom scheme of separation to vary
over a (not very well explained) small class of sets, but then attempt
to expand the universal quantifier in the definition of "uncountable"
to vary of some (not very well explained) larger class of sets.
According to a Platonist, it's true, there's a "real world" of sets,
where the axiom scheme of separation is true and the conclusion is
also true. But it's not necessary to believe that, to believe that the
reasoning is valid. The reasoning is simple first-order logic, and it
holds good as long as you don't confuse yourself by trying to make
the quantifiers in it range over different domains.
|> 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.
|
|I have no idea what you think I may have or would consider being taken
|in by.
You just illustrated my point.
You keep using terms (e.g. "new set") that are pretty vague, and you
seem either not to be aware of it, or to think that it's good enough
for your purposes. My advice is not just some debating ploy. Nobody
is better off by your continuing to make the kinds of errors you've
been making.
Keith Ramsay
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