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

From: J.E. (troubled6man_at_yahoo.com)
Date: 09/30/04 

Date: 30 Sep 2004 08:12:57 -0700

> > > > 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?

> > ... 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.

> > 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.

> 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.

> It's a theory about one set: all of them.

I lost you here. There is no set of all sets.

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