FOM: Dedekind's isomorphism theorem

Charles Parsons parsons2 at
Mon Oct 12 18:19:06 EDT 1998

John Mayberry writes (12 October):

>	There is nothing wrong with the proof that all models of these
>second order axioms are isomorphic (we don't have to talk about the
>"standard structure" here). But for the proof to get off the ground we
>must assume that there is a transfinite set with a power set. Of course
>most mathematicians simply take that assumption for granted. But if you
>don't take it for granted - and clearly Sazonov doesn't - then you are
>left with the problem of how to lay the foundations for natural number

Charles Silver challenges this, in my view rightly, but the point could be
put more sharply. It's misleading to focus on full second-order arithmetic
and its standard models. Dedekind's categoricity theorem can be proved in
second-ortder theories with only first-order comprehension (with
second-order parameters), as in Solomon Feferman and Geoffrey Hellman,
"Predicative foundations for arithmetic", Journal of Philosophical Logic 24
(1995), 1-17, theorem 5. There's no invocation of a power set. To see that
an alternative number sequence is isomorphic to a given one, one needs
induction with reference to each sequence, which of course one doesn't have
in PA, but one doesn't invoke any totality of all putative number sequences
or any other such totality of sets or classes. I used this observation in a
discussion of the philosophical issues in "The uniqueness of the natural
numbers", Iyyun 39 (1990), 13-44.

This is no refutation of Sazonov, since the proof certainly uses devices
that would be unacceptable to a strict finitist.

Charles Parsons

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