There are many interpretations of arithmetic in set theory. The
Zermelo interpretation, for example, begins with the empty set and applies the singleton operator as successor:
$$0=\{\ \}$$
$$1=\{0\}$$
$$2=\{1\}$$
$$3=\{2\}$$
and so on… The von Neumann interpretation, in contrast, is guided by the idea that every number is equal to the set of smaller numbers.
$$0=\{\ \}$$
$$1=\{0\}$$
$$2=\{0,1\}$$
$$3=\{0,1,2\}$$
In each case, one equips the numbers with their arithmetic structure, addition and multiplication, and in this way one arrives at the standard model of arithmetic
$$\langle\mathbb{N},+,\cdot,0,1,<\rangle$$
In those two cases, there isn't much at stake because ZFC proves that they are isomorphic and hence satisfy exactly the same arithmetic assertions. In each case, the interpretations provably satisfy the axioms PA of Peano arithmetic.
But furthermore, each of these interpretations satisfies many additional arithmetic properties strictly exceeding PA, such as Con(PA) and Con(PA+Con(PA)), which are all ZFC theorems.
My main point, however, is that these further properties are not provable in PA itself, if consistent, and so my question is whether there is an interpretation of arithmetic in set theory realizing exactly PA.
Question. Is there an interpretation of arithmetic in set theory, such that the ZFC provable consequences of the interpretation are exactly the PA theorems?
What I want to know is whether we can interpret arithmetic in ZFC in such a way that doesn't carry extra arithmetic consequences from ZFC?
Best Answer
This is equivalent to the $\Sigma_1$-soundness of $\mathsf{ZFC}$ (and this equivalence is highly robust to replacing $\mathsf{PA}$ with some other theory):
If $\mathsf{ZFC}$ is $\Sigma_1$-sound then the answer is yes as follows: consider "the $\alpha$th (in the sense of $<_L$) constructible model of $\mathsf{PA}$, where $2^{\aleph_0}$ is the $\alpha$th cardinal of uncountable cofinality." This definition corresponds to an interpretation $\Phi$ of $\mathsf{PA}$ in $\mathsf{ZFC}$, which is extremely "indecisive:" for any (countable) $\mathcal{M}\models\mathsf{ZFC}$ and any sentence $\theta$ such that $\mathcal{M}\models\mathsf{Con}(\mathsf{PA}+\theta)$, there is a forcing extension $\mathcal{M}[G]$ of $\mathcal{M}$ such that $\Phi^{\mathcal{M}[G]}\models\mathsf{PA}+\theta$. (Just move $2^{\aleph_0}$ far enough up to "grab" a constructible model of $\mathsf{PA}+\theta$, in the sense of $\mathcal{M}$.) Since $\mathsf{ZFC}$ is $\Sigma_1$-sound, as long as $\mathsf{PA}\not\vdash\theta$ there is some countable $\mathcal{M}\models\mathsf{ZFC}$ with (something $\mathcal{M}$ thinks is) a constructible model of $\mathsf{PA}+\neg\theta$.
On the other hand, suppose $\mathsf{ZFC}$ is $\Sigma_1$-unsound. Then there is some Diophantine equation $E$ such that $\mathsf{ZFC}$ thinks $E$ has a solution but $E$ does not in fact have a solution. Now suppose $\Phi$ were an interpretation of $\mathsf{PA}$ into $\mathsf{ZFC}$. $\mathsf{ZFC}$ may not realize that $\Phi$ is in fact an interpretation of full $\mathsf{PA}$, but $\mathsf{ZFC}$ will see that $\Phi$ is an interpretation of at least $\mathsf{I\Sigma_1}$ (since that theory is finitely axiomatized), and since $\mathsf{I\Sigma_1}$ is ($\mathsf{ZFC}$-provably) $\Sigma_1$-complete we will have that $\mathsf{ZFC}$ thinks that $\Phi$ thinks that $E$ has a solution. So "$E$ has a solution" is a non-theorem of $\mathsf{PA}$ (assuming $\mathsf{PA}$ is $\Sigma_1$-sound of course, but I think that's fair), which is provable in every interpretation of $\mathsf{PA}$ into $\mathsf{ZFC}$.