Re: A question on GIT.

From: Andrew Boucher (Helene.Boucher_at_wanadoo.fr)
Date: 09/10/04 

Date: 9 Sep 2004 23:01:10 -0700

Andrew Boucher <Helene.Boucher@toto.fr> wrote in message news:<Helene.Boucher-9D86CB.22241209092004@news.wanadoo.fr>...
> Could the proof go through in some other way, without the successor
> axiom? Well, no. Without the successor axiom one can only model a
> theory by the theory itself or something smaller than the theory,
> because again you can't prove that bigger things exist. But
> second-order PA has a finite number of finite-length axioms.
> (Second-order here means a two-typed system, with big- and little-
> letters, so it still can be considered "first-order" logic.) So one
> could not prove a model, which would have to be infinite, exists.
> (Notice the difference with first-order PA, which has induction as a
> schema. The theory then has an infinite number of axioms, so an
> infinite number of things, and one could then model the theory of
> first-order PA by itself.)

On mature reflection, bad argument! For one thing, in order to be a
counter-example to the Completeness Theorem, second-order PA would
need to be consistent, which I can't assume. So I guess the answer is
"no", I don't have a knock-down argument that the completeness theorem
cannot be proved in (PA2 - Successor Axiom). The proof that is used
to establish the completeness theorem does not work (see my comments
about Lindenbaum's Lemma). But there might be a workaround. I doubt
it, but it order to concretize my doubts I would need to present a
consistent theory without a model, and I don't think I'll be able to
do that. Sorry!

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