Re: Skolem's Paradox and why is math the way it is?

From: Ross A. Finlayson (raf_at_tiki-lounge.com)
Date: 09/30/04 

Date: Thu, 30 Sep 2004 13:45:58 -0700

Hi,

Read this:

http://www.cs.nyu.edu/pipermail/fom/2004-August/008489.html

Does Goedel submarine higher-order logic?

"J.E." wrote:

> > > > > Why do we care about "uncountably many" real numbers when in reality
> > > > > there are only a countable number that we can "prove theorems about"?
> > > >
> > > > Actually, that's not true. Although there can be only a countable
> > > > number of theorems, they can still address an uncountable number of
> > > > reals. As a simple example, there are an uncountable number of reals
> > > > in the Cantor set, and a simple procedure by which to determine given
> > > > any real if it is a member (namely, can its decimal expansion, when
> > > > convereted to base 3, be written without using the digit 1).
> > > >
> > > > Jonathan Hoyle
> > >
> > > Thank you for your polite response. I do think that I need to ask my
> > > question more clearly. There are many sets, we agree on that. Some
> > > sets have bijections with the set of naturals, others don't, we agree
> > > on that.
> >
> > I don't. You can apply a canonical ordering operator onto any set that is
> > not ordering-sensitive.
>
> Which statement were you disagreeing with? Are you saying that all
> sets have bijections with the set of naturals, because all I said was
> that some do and others don't?
>

No, only infinite sets are equivalent, in my theory. There are also
alternatives where that is not so.

This has to do with defining, or rather, acknowledging, that P(X) = X + 1.

>
> > > ... But each time you use an axiom to prove a theorem, the axiom
> > > produces, at most, a finite number of sets, and you can use each axiom
> > > only a finite number of times in a single proof. Doesn't it seem that
> > > each proof generates only a finite number of sets, even though some of
> > > these sets might be considered "large" themselves? And if the number
> > > of theorems is "somewhat countable", then doesn't it seem like there
> > > is only a "somewhat countable" number of sets that we can prove
> > > theorems about?
> > >
> >
> > I eschew axioms, but if you call infinity an axiom then it generates
> > infinitely many sets and all the proofs about those sets.
>
> The axiom of infinite I know asserts the existance of ONE set (that
> either is, or at least contains omega). And then the axiom of
> specification creates a countable number of subsets, and one can
> invoke the other axioms a finite number of times, and there is still
> just an countable number of sets you've made.
>

I just say there is at least nothing and then that through excluded middle there
is everything else, and that the axiom is actually a theorem.

In the above reference, there is discussion about Goedel essentially agreeing
with Skolem and Loewnheim. Also, the reals are the reals are the reals.

>
> > > You, for instance, cited a theorem about ONE set (the cantor set). So
> > > there is at least one set in the universe. But you want be to believe
> > > that you just proved a theorem about many sets. However, all you
> > > proved is that there is a certain subset of another set, namely the
> > > cantor set is a subset of the reals. That is a theorem about ONE set.
> > > How am I supposed to tell if there are as MANY sets in the universe
> > > as you seem to want me to believe there are? Thank you again.
> >
> > Easily, put them in a line.
>
> The axioms don't put anything into a line, the whole point is that the
> axioms don't generate ALL the points on the line, only a countable
> number of them. Consistently one can add more axioms to have more
> numbers on the line, that's what the diagonal arguement says, but if
> one doesn't add more axioms, then you only have a countable number of
> points that the original ZF(C) axioms talk about directly. And sadly,
> even after adding more axioms you still only have a countable number.
>

That's not sad. There are just "countably" many, infinite sets are equivalent.

>
> > Are second order logic and "countable" models of the "uncountable" not
> > slippery slopes? (They are.)
>
> I don't see how it is a slippery slope to be clear what you are
> tlaking about and what you are not.
>

The issue I hope to make clear to you is that pushing off the resolution into
high-order logics, in an admitted way that there is no resolution in resort to
some metalogic, is handwaving that does not offer a resolution, and that if a
resolution is to exist it exists in first order logic, and that all high-order
and meta logic is enframed or emposed within first order logic.

>
> > It's a theory about one set: all of them.
>
> I lost you here. There is no set of all sets.

Why not?

Warm regards,

Ross F.

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