Re: GCH vs. Axiom of Choice.

From: J.E. (troubled6man_at_yahoo.com)
Date: 12/24/04 

Date: 24 Dec 2004 15:09:18 -0800

KRamsay wrote:
> 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?

Consider the following axioms:

A1-EmptySet: EyAx ~xey
A2-Equality: AxAy( x=y <=> Az( zex <=> zey) )
A3-Pairing: AxAyEzAt( tez <=> (t=x or t=y))
A4-Union: AxEyAz( zey <=> Et( zet & tex) )
A5-Powers: AxEyAz( zey <=>At( tez => tex) )

Now P(n) for the finite ordinals n, is just what anyone would expect,
since all the subsets can be built up from union and pairing. So let's
add infinity.

A6-Winfinity: EW (W contains all f.o.)
A7-Ninfinity: ENAn( neN <=> AW((W contains all f.o.)=>neW)

Now if N is that smallest set and itself an ordinal (which I still
think is unclear at this point at least), then P(N)={X:En (neN &
XeP(n))}, the set of all finite subsets. So since in ZFC there are
more subsets than just the finite ones, these more subsets that belong
to P(N) must owe their existence to the separation or replacement
axiom(s).

So if P(N) has more elements than {X: En (neN & XeP(n))} has, then
these extra elements come not from some weird reinterpretation of
"e", but from a more liberal definition of set, one that allows
more things to be sets.

Q. So what exactly will be allowed to be a set that wasn't before?
A. Some objects that contain infinite subcollections of the infinite
sets.

Q. Which such objects?
A. The objects Y such that there is a formula S[a] and an prexisting
infinite set X, such that "Ay( yeY <=> (yeX & S[y]) )" is true.

Q. How many of them are there?
A. At most one per goedel number per set X.

So all these new sets Y where X=N, should belong to P(N), but that's
still just as many as there are goedel numbers, in fact the finite
subsets are taken care of too with the appropriate choice of S[a]. Are
there more sets in P(N)? I don't see any indication that there are,
just like there was no indication that the infinite subsets were in
P(N) in the axiom system A1-A5.

A Planontist might insist that (1) there are more subsets in P(N) right
now, as strongly as that (2) there were infinite subsets in P(N)
*always*, all that we lacked was a method to prove it, but what are the
grounds for that platonic assertion? Basically that it was unfair to
assume that the objects in question weren't sets, just because we
couldn't prove that they were. Is that a standard that can be upheld
consistently? I don't know, but I do know that it'd be unusual to be
consistent about that standard, since ZFC doesn't follow that standard
when the Separation axiom is introduced, since that axiom excludes the
object that contains all the sets. Why should the set of all sets
exist? Well, if the axioms A1-A7 are consistent, then so are the
axioms B1-B7

B1-M.EmptySet: EyAx xey
B2-M.Equality: AxAy( x=y <=> Az( ~zex <=> ~zey) )
B3-M.Pairing: AxAyEzAt( ~tez <=> (t=x or t=y))
B4-M.Union: AxEyAz( ~zey <=> Et( ~zet & ~tex) )
B5-M.Powers: AxEyAz( ~zey <=>At( ~tez => ~tex) )
B6-M.Winfinity: EW (W m.contains all m.f.o.)
B7-M.Ninfinity: ENAn( ~neN <=> AW((W m.contains all m.f.o.)=>~neW)

Which are obtained by putting a not on every atomic formula in every
axiom, proof, theorem, and definition of ZFC. And with those axioms
there is an element that contains everything, that's that first axiom,
M.EmptySet. And M.Separation makes m.sets bigger, much like Separation
makes sets smaller. So if platonists want to play the "include as
many sets as possible game", then the m.sets are at least as good as
the small sets, so they should either both exist or both not exist,
which is clearly incompatible with the axiom of separation, which
platonically would then be a FALSE axiom.

I'm personally willing to consider a universe where big sets and little
set coexist in harmony. If we want to play the "allow whatever
objects can be sets be sets game", but if we are playing the "some
things are sets and others aren't game", then I find the insistence
that the power set "contains them all[sic]", just hyprocritical
hype/fluff, when "all" is mocked up as being "all the
possible", when it's really just "all the popular sets", because
then this becomes sociology and not math.

> |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.

> 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. 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.

> 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. That's where the whole
arguement becomes viciously circular. 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.

> 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. Some predicates capture sets "~xex & xex" captures the empty
set (if we use ZFC), "~xex or xex" captures the m.empty m.set (if we
used m.ZFC), the syntax "~xex" captures no set in any axiom system,
because in general for a set y the formula S[x] captures the set y if
and only if Ax( xey <=> S[x]). And so using ~xex for S[x], we get "Ay(
(y is captured by ~xex) <=> Ax( xey <=> ~xex) )", but if Ax is
interpreted the standard way, then Ax( xey <=> ~xex) => (yey <=> ~yey),
and since the latter is false, so must the former, so we can conclude
"Ay it is not the case that y is capture by ~xex". Now what remains to
be shown is if there are sets that are not cpatured by a formula. That
would be something. The diagonal subset of naturals would, I gather,
be your proposal, even though the axiom of separation doesn't actually
prove that it exists, but if the separation scheme excludes valid sets
like m.0 and m.1 and m.N, and doesn't allow sets you like like the
diagonal subset of naturals, then it just seems like a bad axiom
scheme.

> | 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".

A1-A7 can have a different power set than A1-A7+separation. So it
depends on the other axioms, and separation is a terrible axiom.

> |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.

Powerset is consistent with saying that no infinite subsets of the
naturals exist. The problem is figuring out what is a set.

> 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.

What about the set of all sets?

> | 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.

I was saying that from within ZFC you can't prove that ZFC has more
sets in it than the sets definable in the ZFC language.

> |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.

If someone says that the powerset isn't countable, I have two choices,
that the set of functions available for counting is too small to count
it (like in my example of V=N+2), or that the power set contains more
elements than finite ordinals. It seems like either the former is
true, or that we need a stronger language than ZFC to say what is
actually in ZFC. And if we want ZFC to have as much as possible
(instead of having it JUST have those things definable in ZFC), then I
don't know why we exclude the object that contains all of the sets, for
starters.

As for bannanas, maybe it comes down to predicativity. If someone
defined all the sets and then pointed at one and said "hey that one P
contains all the subsets of that one X", then that's be fine, but we
are talking about using the definition of set to create a new set.
It's totally different then discussing bananas. Bananas don't contain
other bananas. Either every set has a sentence that describes it, or
it doesn't. Either there are more sets than sentences, or their
aren't. Either there is no set that contains all sets, or there is.

> | 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".

If V=NU{X} and "Ax xeX", then X itself would be a finite ordinal, since
it belongs to the smallest (only) set that contains all the finite
ordinals. It clearly depends on the other axioms. Now the problem is
that if separation is supposed to be able to shave off all the things
that aren't the finite ordinals, then if we have sets that aren't
defined by ZFC, for instance that diagonal subset of naturals, then HOW
does separation exclude it, so HOW do we know that it isn't "a finite
ordinal"? Are we eating the cake, or are we having our cake?

> | 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?

I read some nonstandard analysis 7 years ago. What I've read recently
is the proof of Skolem's theorem, since everyone on Usenet seemed to
use different definitions, that wasn't going so well except to annoy
people.

Relevant Pages

  • Re: GCH vs. Axiom of Choice.
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  • Re: Resolving the paradoxes of set theory
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