elementary equivalence and incompleteness

Question

I read the following line in a text on set theory:

"Peano Arithmetic has continuum many non-isomorphic countable models (including the standard model omega), all of them elementary equivalent."

There are countable models of PA + ~Con(PA). Since PA + Con(PA) is true in the standard model, why is the quoted statement not false?

Thanks.

Answer 1

The quote is somewhat ambiguously written - it would be better written as:

There is a set $\{\mathcal{N}_i: i\in2^{\aleph_0}\}$ of continuum-many pairwise-nonisomorphic countable models of PA, including the standard model, which are elementarily equivalent.

These aren't all the countable models of PA, even up to isomorphism, of course.

The point of the quoted statement is that, while there are indeed nonstandard models which differ from the standard one at the level of theories, there are also nonstandard models - lots of them, even! - which look exactly the same as far as the theories are concerned. For instance, there are nonstandard models of PA which also think that PA is consistent; "Con(PA)" does not characterize just the standard model.

There are also lots of models which don't look the same, at the level of theories; it's a good exercise to show that there is a set $\{\mathcal{M}_i: i\in 2^{\aleph_0}\}$ of continuum-many countable models which are pairwise elementarily non-equivalent, that is, $$i\not=j\implies Th(\mathcal{M}_i)\not=Th(\mathcal{M}_j).$$ But that's a different phenomenon than the one being talked about.