# Does ZF(C) prove countable reflection?

Is the following a theorem of $$\sf ZF(C)$$?

Countable reflection: If $$\phi$$ is a sentence in which $$W$$ is not free, then:

$$\phi \to \exists W: |W|= \omega \land \operatorname {Transitive}(W) \land \phi^W$$

Where $$\phi^W$$ is the sentence obtained by merely bounding all quantifiers in $$\phi$$ by $$\in W$$.

That is: every true sentence is reflected upon a countable transitive set.

So all true sentences are reflected upon some elements of $$V_{\omega_1}$$ (or $$V_{\omega_2}$$ in case of $$\sf ZF$$).

• Yes, this follows from the reflection theorem and Lowenheim-Skolem. Jan 22 at 22:40
• The question twice mentions "formula" (once with the constraint that $W$ isn't free) and twice mentions "sentence(s)". I assume (and Monroe Eskew and Ali Enayat apparently also assumed) that you mean "sentence", because the version with "formula fails --- the formula $(\exists y)\ y=x$ would be a counterexample. Jan 23 at 4:53
• @AndreasBlass, Agreed! Thanks for the correction. I've edited the question. Jan 23 at 9:22

As Monroe Eskew points out in his comment to the question, the positive answer is well-known for ZFC, thanks to the ZF reflection theorem and the Löwenheim-Skolem Theorem (in the form: every model in a countable language has a countable elementary submodel). See, e.g., part (iii) of Theorem 12.14 in Jech's canonical textbook Set theory, third millenial edition.

Note that ZFC can be weakened in the above to ZF+ DC (DC = dependent choice) since it is well-known that in the presence of ZF, DC implies (and indeed is equivalent to, see, e.g., Asaf Karagila's note) the statement that every structure in a countable language has a countable elementary submodel.

The point of this answer is to point out that even DC can be eliminated from the proof of countable reflection.

Theorem. ZF proves every instance of countable reflection.

Proof: In light of the fact that the Mostowski collapses can be carried out in ZF, and the constructible universe L satsifies ZFC (and therefore L satisfies "every model in a countable language has a countable elementary submodel), the proof is complete once we note:

$$(*)$$ For any set-theoretical sentence $$\phi$$, if $$\phi$$ has a transitive model, then the sentence "$$\phi$$ has a transtive model" holds in the constructible universe L.

The above statement, as far as I know, was first noted in a (famous) paper of Barwise and Fisher entitled "The Shoenfield Absoluteness Lemma" (Israel J. Math. 8 1970, pp.329-339) in which a fine-tuning of the Shoenfield Absoluteness Lemma is presented (in particular, it is shown that the Lemma can be proved without invoking DC).

More specifically, $$(*)$$ follows from Theorem 1b (page 336) of the Barwise-Fisher paper, which states that if $$\phi$$ has a transitive model $$A$$ with $$\rho(A)=\alpha$$ (where $$\rho$$ is the usual ordinal-valued rank function on sets), then there is a transitive model of $$\phi$$ in the $$\kappa^{+}(\alpha)$$-th level $$L_{\kappa^{+}(\alpha)}$$ of the constructible universe $$L$$, where $$\kappa^{+}(\alpha)$$ is the next admissible ordinal after the least admissible ordinal above $$\alpha$$.