3
$\begingroup$

Let's speak of the theory $\sf ZC + rank$ as the first order set theory with axioms of Extensionality, Separation, infinity, and choice (written as usual), plus iterative powers and foundation, those are:

Iterative Power: $\forall \text { ordinal } \alpha \exists x : x=P^\alpha(\emptyset)$

where $P^\alpha(\emptyset) = \bigcup \{P(P^\beta(\emptyset)) : \beta < \alpha\} $

Foundation: $\forall x \exists \text { ordinal } \alpha: x \in P^\alpha (\emptyset)$

Now ordinal is meant to be a Von Neumann ordinal.

is every extension of $\sf ZFC$ interpretable in a finite extension of $\sf ZC + rank$?

is every finite extension of ZFC interpretable in a finite extension of ZC + rank?

$\endgroup$
5
  • 2
    $\begingroup$ Definitely not every extension - ZFC has uncountably many pairwise incompatible extensions (e.g. for any sequence $(a_n)\in\{1,2\}^\omega$ consider ZFC+"$2^{\aleph_n}=\aleph_{n+a_n}$" for all $n$), but ZC+rank has only countably many finite extensions, and each interprets only finitely many theories. $\endgroup$
    – Wojowu
    Jun 8, 2020 at 16:51
  • $\begingroup$ @Wojowu, then it might be more plausible to ask if every finite extension of ZFC is interpretable in a finite extension of ZC + rank? $\endgroup$ Jun 8, 2020 at 18:30
  • $\begingroup$ @Wojowu, I just want to ask about the theories "ZFC + for all $n:2^{\aleph_n}=\aleph_{n+a_n}$" now each of these must be "effectively generated" so it must have a clear definition of its terms, it appears to me that $a$ is a sequence (so its domain is N) of binary strings which would code some reals I suppose. But this sequence must be definable otherwise the theory "ZFC + for all $n:2^{\aleph_n}=\aleph_{n+a_n}$" is not well defined, and so not effectively generated. So we only have countably many such extensions. $\endgroup$ Jun 9, 2020 at 9:38
  • $\begingroup$ @EmilJeřábek, yes I assume consistency and that ZC+rank is $\Sigma_1$-sound. $\endgroup$ Jun 9, 2020 at 9:42
  • $\begingroup$ @Wojowu, I see where I missed, I should have said any recursively axiomatizable extension of ZFC. I noticed that after reading Emil's answer, although my comment preceded this answer and it was clear that I was meaning effectively generated extensions of ZFC. $\endgroup$ Jun 9, 2020 at 12:34

1 Answer 1

6
$\begingroup$

$\let\res\restriction\def\N{\mathbb N}$By the discussion in comments, it is missing from the question that all the theories are assumed consistent (otherwise the answer is trivially yes, as everything is interpretable in the inconsistent extension) and recursively axiomatizable (otherwise the answer is trivially no, as there are $2^\omega$ pairwise incompatible complete extensions of ZFC, while only countably many of these are interpretable in any given consistent extension of ZC + rank).

With these caveats, the answer to both questions is yes, for general reasons that work for most other pairs of natural theories. In what follows, if $T$ is a theory with a fixed r.e. set of axioms, let $\Box_T$ denote the formalized provability predicate for $T$, and $T\res n$ the theory axiomatized by the axioms of $T$ with Gödel number below $n$.

Proposition. Let $S$ be an extension of $I\Delta_0+\mathrm{EXP}$, and $T$ be a recursively axiomatized theory such that $$\tag{$*$}S\vdash\Box_{T\res n}\phi\implies T\vdash\phi$$ for all $n\in\N$ and all $T$-sentences $\phi$. Then every consistent r.e. extension of $T$ is interpretable in a consistent finite extension of $S$.

Observe that $(*)$ holds whenever $S\subseteq T$ (or just $S$ is $\Sigma_1$-conservative over $T$) and $T$ is a locally essentially reflexive theory, for example $S=\mathrm{ZC+rank}$ and $T=\mathrm{ZFC}$. But in fact, $(*)$ holds whenever $S$ is $\Sigma_1$-sound, which pretty much any natural theory is. Also, it is not difficult to show that if $S$ is locally essentially reflexive, then condition $(*)$ is necessary.

Proof. Let $U\supseteq T$ be recursively axiomatized and consistent. Then $(*)$ implies that $$V=S+\{\mathrm{Con}_{U\res n}:n\in\N\}\equiv S+\{\neg\Box_{T\res n}\neg\phi:n\in\N,\phi\in U\}$$ is consistent; since it is an extension of $S$ by a r.e. set of $\Pi_1$-sentences, there exists a $\Pi_1$-sentence $\psi$ such that $S+\psi$ is a $\Sigma_1$-conservative extension of $V$ by a theorem of Lindström [1]. In particular, $S+\psi$ is consistent, and since it proves $\mathrm{Con}_{U\res n}$ for all $n\in\N$, it interprets $U$ by the usual interpretation existence lemma. QED

Reference:

[1] Per Lindström: On partially conservative sentences and interpretability, Proceedings of the American Mathematical Society 91 (1984), no. 3, pp. 436–443, doi: 10.2307/2045318.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.