Re: Resolving the paradoxes of set theory

From: George Greene (greeneg_at_cs.unc.edu)
Date: 11/02/04 

Date: 2 Nov 2004 14:57:01 -0800

daryl@atc-nycorp.com (Daryl McCullough) wrote in message news:<cm32ud0pce@drn.newsguy.com>...
> George Greene says...
> >
> >Norman Megill <nm@nospam.see.signature.domain.invalid> wrote
>
> >> The proof that there is no surjection from a set to its power
> >> class does not require the power set axiom.
> >
> >Of course it does.
>
> No, it doesn't.

Of course it does. Powersets are not guaranteed to exist.

> The use of P(x) is just a notational convenience.

That's how he MEANS it. But even THAT is NOT sufficient
to guarantee that powersets exist. It matters a lot just
WHAT string, in the original language, P(x) is intended to
abbreviate. He can't clarify this without invoking CLASSES
AS OPPOSED to sets. He can say that P... is the type of thing
that has a class expansion that in turn has an expansion as
something that has to be a phi()-like syntactic formula.
But that will be about a powerCLASS as OPPOSED to a powerSET.

> State the theorem this way:
>
> forall x, forall f, if f is a function whose domain is x,
> then there exists a set x' such that x' is a subset of x
> and x' is not in the range of f

This is not even the right theorem.
What is a function? Is it a SET of ordered pairs, or
can it be a class of them? What is x? Is it a set or
a class? ALL OF THIS MATTERS. The "set x'" that you talk
about is in fact NOT guaranteed to BE a set (as opposed to
a class) unless x and f are too. More to the point, the
fact that the word "power" doesn't even occur in this
makes it important-all-over-again to clarify just WHAT
P() is EXPANDING to, "as a notational convenience".
"The powerset axiom" whose fate we are debating here is
ITSELF nothing but "a notational convenience".

> The proof uses separation (given a set x and a formula Phi(y),
> there is a set x' consisting of all y in x satisfying Phi(y))
> but it doesn't use the power set axiom.

As you have incorrectly stated your version, there is no set
AVAILABLE for the separation axiom to use. JP is operating
in a world where classes exist, or where he is at least free
to talk as though they did. If he is not going to concede their
actual existence then he is not even in a first-order language;
Every occurrence of a class functor has to be translated down
into something else, something formulaic. This is NOT going to
automatically entail existence of any terms. If such existence is
exactly what you are trying to prove then the translation is simply
not going to be right.

What you were TRYING to say was
"If f is a CLASS of ordered pairs and x (which is also
a CLASS) = Dom(f), and y (which is also a CLASS) = Rng(f),
then there is subCLASS x' of x that does not belong to y".

>From this, the ACTUAL theorem, which is the special case
of this when x is a set -- whence it follows that f and y
are as well -- is a corollary. But it also requires some
set-existence axioms; I think what is happening there
is that you just get to use Replacement as the only set
existence axiom. If we are going to start talking about
redundant axioms, then Separation is redundant as well.

that is not an element of Rng(f) (which is also
subCLASS x of Dom(f) (Dom(f) is also a class) f

Relevant Pages

  • Re: Resolving the paradoxes of set theory
    ... It is a very deep semantic paradox. ... > This has absolutely nothing to do with powersets ... > It's equally inconsistent with the DENIAL of regularity, ... > stop blaming everything on the powerset axiom. ...
    (sci.logic)
  • Re: Grothendieck Universes
    ... My immediate guess is that it seems to guarantee no more ... The axiom of universes implies that there exists a proper class of ...
    (sci.math)
  • Re: An uncountable countable set
    ... Virgil schrieb: ... But it does not guarantee what is existing there. ... I do not know what "axiom of infinity", 'mueckenh" is referring to, but ...
    (sci.math)
  • Re: An uncountable countable set
    ... But it does not guarantee what is existing there. ... I do not know what "axiom of infinity", 'mueckenh" is referring to, but ... "naturals" without having to count anything. ... Why does "mueckenh" think any natural numbers are not numbers? ...
    (sci.math)