FOM: Two questions
Charles Parsons
parsons2 at fas.harvard.edu
Wed Jan 26 18:04:10 EST 2000
I don't find what Mark Steiner says about what Boolos and others call
"Frege's Theorem" very clear. Let me try to state the matter in my own
terms.
Frege is credited with proving that if one augments second-order logic with
full comprehension and a cardinality operator NxFx and the axiom:
(HP) NxFx = NxGx iff there is a one-one mapping of the Fs onto the Gs,
then second-order arithmetic is interpretable in the resulting theory. In
particular every number (in fact every cardinal, i. e. every y that is NxFx
for some F) has a successor.
I leave it to others to say whether this is "core mathematics". The beauty
of this formulation is that it offers something independent of Frege's
explicit definition of cardinals as "equivalence classes" and the resulting
reliance on the disastrous axiom V.
Of course it follows that in a model of this theory, the domain has to be
infinite. I think that's what creates the appearance of getting something
for nothing. Assuming the mapping NxFx from classes or "concepts" into
individuals is not so innocent as it seems. (See Boolos, "The standard of
equality of numbers," in his collection _Logic, Logic, and Logic_ (Harvard
UP, 1998).)
I don't think Mark has quite right what Boolos and Richard Heck showed
about the proofs. It was that the sketch in Grundlagen is of an incorrect
proof. The proof in Grundgesetze is correct but is not the same argument as
that sketched in Grundlagen. (See Boolos and Heck, "Die Grundlagen der
Arithmetik, §§82-83," in Boolos's collection.)
But now what's the problem about ZFC? In the form in which I formulated it,
of course the theorem can be proved in ZFC. (I don't think Mark thinks
otherwise.) But the question might be different, whether we can define in
the language of ZF a cardinality operator and prove (HP). If F and G have
sets as extensions, then we can define NxFx even without Choice by the
well-known method of Scott. But without the restriction to setlike Fs, it
can't even be stated in the language of ZF. Even in the second-order
language of ZF, I don't see how to avoid such a restriction unless one
makes an assumption like von Neumann's, that F has a set as extension
unless it is equinumerous with the universe, so that there is only one
"cardinality" of proper classes.
I would hesitate to say that the assertion that there is such a cardinality
operator, not restricted to sets, belongs to core mathematics. But I'd be
interested to know what others say about this.
Charles Parsons
More information about the FOM
mailing list