# Can we interpret arithmetic in set theory, with exactly PA as the ZFC provable consequences?

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?

• Does $ZFC$ $-$ Infinity inherit the same problems as $ZFC$ even though it is bi-interpretable with $PA$? May 27 at 4:35

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 $$\mathsf{ZFC}$$.

• "It seems you are assuming that that the actual consistency statements are the same as those realized in models of ZFC. Is that right?" I'm not sure what that means. I'm assuming $\mathsf{ZFC}$ is $\Sigma_1$-sound, that's all. May 26 at 1:14
• Not sure why the meaning isn't clear. The answer seems to be: Yes, provided ZFC is Sigma_1 sound, otherwise not. If you edit to make this clear, I'll accept. May 26 at 1:46
• @JoelDavidHamkins To be fair, I do think the interpretation I've used in the $\Sigma_1$-sound case is silly. But I've changed that to "artificial." (And I've fixed the $\mathsf{I\Sigma_1}$ thing.) May 26 at 2:03
• @JoelDavidHamkins Fair enough, I'll delete it. I do think there is an interesting follow-up question re: identifying a better behaved class of interpretations, though. May 26 at 2:12
• @JoelDavidHamkins Aha! This was what the shift to $\mathsf{I\Sigma_1}$ was for; I just borked it and then forgot. Fixed! (I am assuming that $\mathsf{PA}$ is $\Sigma_1$-sound, though.) May 26 at 2:28

The standard terminology is that an interpretation $$I$$ of a theory $$U$$ in a theory $$T$$ is faithful if for all sentences $$\phi$$ in the language of $$U$$, $$T\vdash\phi^I\iff U\vdash\phi.$$ (Here and below, I will assume all theories to be recursively axiomatized.) A classical result of Feferman, Kreisel, and Orey [1] states:

Theorem 1 (Feferman, Kreisel, Orey). Let $$T$$ be a reflexive theory which includes a modicum of arithmetic and which is $$\Sigma_1$$-sound. Then all theories interpretable in $$T$$ are faithfully interpretable in $$T$$.

They also show that conversely, if a $$\Sigma_1$$-sound theory $$U$$ is faithfully interpretable in an essentially reflexive theory $$T$$, then $$T$$ is itself $$\Sigma_1$$-sound.

Corollary: PA is faithfully interpretable in ZFC if and only if ZFC is $$\Sigma_1$$-sound.

Further results on faithful interpretability in reflexive theories were obtained by Lindström [2]; for example, he proved the following characterization:

Theorem 2 (Lindström). Let $$T$$ and $$U$$ be essentially reflexive theories. Then $$U$$ is faithfully interpretable in $$T$$ if and only if $$U$$ is $$\Pi_1$$-conservative over $$T$$ (which is, by itself, equivalent to the interpretability of $$U$$ in $$T$$), and $$T$$ is $$\Sigma_1$$-conservative over $$U$$.

Generalizing Theorem 1 in a different direction, a theory $$T$$ is called trustworthy if all theories that are interpretable in $$T$$ are faithfully interpretable in $$T$$. Thus, Theorem 1 states that $$\Sigma_1$$-sound reflexive theories are trustworthy.

A complete characterization of trustworthiness was given by Visser [3], who in particular proves that the assumption of reflexivity is unnecessary. (Note that the criterion below does not rely on a fixed interpretation of arithmetic, unlike notions such as $$\Sigma_1$$-soundness or reflexivity.)

Theorem 3 (Visser). A theory $$T$$ is trustworthy if and only if it has an extension $$T'$$ (in the same language) such that Robinson’s arithmetic $$Q$$ has a $$\Sigma_1$$-sound interpretation in $$T'$$.

He also derives from this an earlier result attributed to (but apparently unpublished by) H. Friedman:

Corollary. A consistent finitely axiomatized sequential theory is trustworthy.

## References

[1] Solomon Feferman, Georg Kreisel, and Steven Orey: 1-Consistency and faithful interpretations, Archiv für mathematische Logik und Grundlagenforschung 6 (1962), pp. 52–63, doi 10.1007/BF02025806.

[2] Per Lindström: On faithful interpretability, in: Computation and Proof Theory (Börger, Oberschelp, Richter, Schinzel, Thomas, eds.), Lecture Notes in Mathematics vol. 1104, Springer, 1984, doi 10.1007/BFb0099490.

[3] Albert Visser: Faith & falsity, Annals of Pure and Applied Logic 131 (2005), pp. 103–131, doi 10.1016/j.apal.2004.04.008.

• Thank you very much for this extremely informative answer! Perfect. May 26 at 11:06
• You are welcome. May 26 at 11:19
• What is a reflexive theory? May 26 at 13:56
• @AlexKruckman A theory $T$ is reflexive if it proves the consistency of each finite subtheory of $T$. And $T$ is (locally) essentially reflexive if all its extensions (wlog finite) are reflexive, or in other words, if it proves the reflection schema $\Box_{T_0}(\phi)\to\phi$ for all finite subtheories $T_0\subseteq T$ and sentences $\phi$. A useful sufficient condition is that all sequential theories that prove full induction are (even uniformly) essentially reflexive. May 26 at 14:20