# Does 'ZF without powerset' proves consistency of finite pieces of ZF?

Let us have $\text{ZF}^-$ standard set theory without choice or powerset. Let $\text{ZF}$ be the standard set theory without choice. Is the following true?

For any sentence $\theta$ s.t. $\text{ZF} \models \theta$ $$\text{ZF}^- \vdash \exists M \ (M \text{ transitive}) \wedge (M, \in) \models \theta$$

Also, a seemingly stronger question

For any sentence $\theta$ s.t. $\text{ZF} \models \theta$ $$\text{ZF}^- \vdash \exists M \ \exists E \ (M, E) \models \theta$$

It seems to me that that this should not be true, with $L_{\omega_1}$ or $H(\omega_1)$ being a counterexample. Also, I am not sure whether first question is equivalent to the second one - second one allows nonstandard models, but would that make a big difference?

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The answer to both questions is no.

First, let me point out that the phrase "ZF without powerset" is actually more ambiguous than it may appear, in light of the results of the following paper, whose abstract I quote (please follow the link through to the arxiv for the paper). The basic situation is that the replacement axiom is no longer equivalent to the collection and separation axioms when one lacks the power set axiom. Furthermore, the various standard formuations of the axiom of choice are no longer equivalent without the power set axiom.

"What is the theory ZFC without power set?", by Victoria Gitman, Joel David Hamkins and Thomas Johnstone, forthcoming.

Abstract. We show that the theory ZFC-, consisting of the usual axioms of ZFC but with the power set axiom removed---specifically axiomatized by extensionality, foundation, pairing, union, infinity, separation, replacement and the assertion that every set can be well-ordered---is weaker than commonly supposed and is inadequate to establish several basic facts often desired in its context. For example, there are models of ZFC- in which $\omega_1$ is singular, in which every set of reals is countable, yet $\omega_1$ exists, in which there are sets of reals of every size $\aleph_n$, but none of size $\aleph_\omega$, and therefore, in which the collection axiom scheme fails; there are models of ZFC- for which the Los theorem fails, even when the ultrapower is well-founded and the measure exists inside the model; there are models of ZFC- for which the Gaifman theorem fails, in that there is an embedding $j:M\to N$ of ZFC- models that is $\Sigma_1$-elementary and cofinal, but not elementary; there are elementary embeddings $j:M\to N$ of ZFC- models whose cofinal restriction $j:M\to\bigcup j''M$ is not elementary. Moreover, the collection of formulas that are provably equivalent in ZFC- to a $\Sigma_1$-formula or a $\Pi_1$-formula is not closed under bounded quantification. Nevertheless, these deficits of ZFC- are completely repaired by strengthening it to the theory $\text{ZFC}^-$, obtained by using collection rather than replacement in the axiomatization above. These results extend prior work of Zarach.

As we discuss on page 9, it is an open question whether $\text{ZFC}^-$ proves the reflection theorem scheme, in the form: for every formula $\varphi$ and parameter $a$, there is a transitive set $A$ containing $a$ such that $\varphi$ is absolute between $V$ and $A$. It is of course the reflection theorem that one would want to appeal to, in order to prove even that $\text{ZF}^-$ proves the consistency even of the finite pieces of $\text{ZF}^-$, let alone ZF itself. Nevertheless, theorem 5 of the paper establishes that the weaker version of ZFC- does not prove the reflection scheme, and this amounts to a negative answer to your first question for that version of ZF-. The moral of the paper is that one should mean what we call $\text{ZF}^-$, usiing collection and separation rather than replacement.

Now let's consider your questions for that theory. Your first question is actually much stronger than the title question, since you require that the model is transitive. For this reason, the answer is no. For example, ZF proves the sentence $\theta$ asserting that there is a transitive model of $\text{ZF}^-$. But if $M$ is an $\in$-minimal transitive model of $\text{ZF}^-$, then it will satisfy $\text{ZF}^-$, but it will have inside it no transitive model of $\theta$, since then it would have a also a transitive model of $\text{ZF}^-$ inside it, contradicting its minimality.

You say that your second question is "seemingly stronger", but since you have dropped the transitivity requirement, the requirement there in fact is much weaker. But the answer here is also negative, since we can let $\theta$ be the arithmetic assertion $\text{Con}(\text{ZF}^-)$. The theory ZF proves the consistency of $\text{ZF}^-$, but this is not provable in $\text{ZF}^-$ itself, by the incompleteness theorem, and the existence of a model of $\theta$ would imply $\theta$, since any proof of a contradiction in the ambient theory would be correspond to such a proof inside $(M,E)$. So the answer to this question also is no.

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Thanks! I thought there is pretty straightforward way to answer this, but couldn't quite put my finger on it. And the way I was thinking of solving it was much harder than the solution you gave. – Anton Jan 25 '12 at 19:07