Is there a first order formula that is satisfied by $\mathbb{N}$, but not by any other models of Peano axioms?

Question

Is there a first order formula that is satisfied by $\mathbb{N}$, but not by any other models of Peano axioms? For example, is there a formula that expresses "there is an element that is greater than any natural number"? My first try was $$\exists v_0 (0<v_0 \wedge \forall (v_1\; v_1<v_0 \Rightarrow Sv_1 <v_0)). $$ But I see that this formula is not true in non-standard models either, because every non-zero element has a predecessor.

Answer 1

You can appeal to Gödel's incompleteness theorem: clearly the Peano axioms plus your one axiom are still recursively enumerable, and clearly they are strong enough to encode arithmetic (since they contain the Peano axioms). Thus the theory must be incomplete, and in particular have more than one model by Gödel's completeness theorem.

Answer 2

No.

Take any ultrapower of $\Bbb N$ by a free ultrafilter and you have a non-standard model which is elementarily equivalent to $\Bbb N$. If you want a countable model, simply take a countable elementary submodel of that ultrapower defined by some non-standard integer.

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Alternatively, note that $\operatorname{Th}(\Bbb N)$ is a first-order theory, and since it has a pointwise definable model (i.e. a model in which every object is definable, in this case $\Bbb N$ is that model where every object is definable by $S\dots S0$ for the appropriate amount of $S$s), it is not categorical in that cardinality. In other words, it has more countable models. Or, again, if cardinality is not important, just note that it is a first-order theory extending $\sf PA$, and since it has an infinite model, it has one of any cardinality.