[FOM] second-order ZF
Harvey Friedman
friedman at math.ohio-state.edu
Sat Dec 20 16:02:34 EST 2003
On 12/17/03 5:45 PM, "Robert Black" <Mongre at gmx.de> wrote:
> The following question only makes sense if you think (or are willing
> to pretend to think) that we have an unambiguous understanding of
> second-order logic in the sense that given an infinite set S, the
> notion of *every possible* subset of S has an absolute reference (and
> not just a reference relative to a particular model of first-order
> ZFC).
...
> Perhaps first-order ZFC is consistent, but second-order ZF is
> unsatisfiable.
If the last paragraph above is the question, then the question makes
perfectly good sense as a sentence in the language of ZFC, and so we can ask
what its status is in ZFC.
ZFC + Con(ZFC) + "no second order model of ZF"
is consistent if and only if
ZF + Con(ZF)
is consistent.
Furthermore, this equivalence can be proved in extremely weak fragments of
Peano Arithmetic.
Harvey Friedman
More information about the FOM
mailing list