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For some model of ZFC, $M$, we will let $Z_M(H)$ be the following 1-type over $M$:

  • "$H$ is a finite set."
  • "$S \in H$" for all $S \in M$

Then for some model $M'$ of ZFC with a distinguished element $H$, we say that $B(M')$ if $M'$ realizes $Z_M(H)$ over some elementary submodel $M$. The language of $M'$ is $(\in, H)$.

Is if there is theory $T$ in the language $(\in, H)$ such that $\forall M'. B(M') \iff M' \models T$ (assuming ZFC is consistent)?

So, if $T$ exists, it will equal the set of statements that are true in each $M'$. My question then just reduces to (the negation of) if there is some model that that models this set of statements but does not satisfy $B$.

If $T$ exists, we know a couple of things it proves:

  • $T \vdash ZFC$
  • $T$ proves $H$ is a finite set.
  • If ZFC proves that some formula defines a set, then $T$ proves that set is in $H$.
Christopher King
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  • Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on [meta], or in [chat]. Comments continuing discussion may be removed. – Shaun Feb 19 '25 at 12:35

1 Answers1

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Belatedly, it occurs to me that this actually has a simple negative answer. Incidentally, I'll say "$N$ is subfinitizing" in place of the OP's "$B(N)$."

First, motivated by the lengthy comment thread above let me give a detailed explanation of why the phenomenon in the OP does happen. For $M\preccurlyeq N$ models of $\mathsf{ZFC}$, say that $N$ subfinitizes $M$ iff there is some $a\in N$ such that $N\models$ "$a$ is finite" and for each $m\in M$ we have $N\models m\in a$. The OP's property "$B(N)$" corresponds to "$N$ subfinitizes an elementary substructure of itself;" this motivates my alternate terminology "subfinitizing" above.

Let $M$ be any model of $\mathsf{ZFC}$. The following partial $1$-type in the variable $x$ is finitely consistent: $$\{x\mbox{ is finite}\}\cup\{m\in x: m\in M\},$$ and so there is an elementary extension $M'\succcurlyeq M$ in which this type is realized by some element $a$. Stealing a term from constructive logic, every model of $\mathsf{ZFC}$ is subfinite in some elementary extension. On the other hand, if $N$ subfinitizes $M$ then $\omega^N$ must contain an element $N$ thinks is greater than every element of $\omega^M$ (this is essentially the source of the confusion in the comment thread). In particular, this means that no $\omega$-model of $\mathsf{ZFC}$ can be subfinitizing.

Now on to the answer. First, let's consider the case of a theory in the language $\{\in\}$. If $N\models\mathsf{ZFC}$ is countably saturated, then it is subfinitizing. This means that every model of $\mathsf{ZFC}$ is elementarily equivalent to a subfinitizing model, and so the existence of non-subfinitizing models (e.g. $\omega$-models) gives a negative answer to your question.

(Why is every countably saturated model subfinitizing? Well, suppose $N$ is countably saturated. By downward Lowenheim-Skolem it has a countable elementary submodel $M\preccurlyeq N$. By countable saturation, every type over $M$ is realized in $N$ ... including the type you're asking about.)

OK, now what about the richer language? Well, that doesn't change anything: adding a constant symbol doesn't really affect saturation.

Noah Schweber
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