Last night I was thinking about some related statements which follow from ZF+DC, but it actually seems they only need DC to hold in some outer model of the universe. In particular, let $M \models ZF.$ Consider the following claims (all relativized to $M$):


1. For any sentence $\sigma$ such that $M \models \sigma,$ there is a countable transitive model $M\_0 \in M$ such that $M\_0 \models \sigma.$
2. $L^M$ and $M$ agree on $\Sigma\_1$ sentences (Levy's version of Shoenfield absoluteness).
3. $HC^M \prec\_1 M.$


I believe these follow from there being some outer model $N \supset M$ where $N \models ZF + DC,$ or even $M$ existing in some ambient universe $V$ such that $V \models ZF+DC \wedge ``M \text{ is transitive}" \wedge \text{ } \omega\_1 \subset M.$ E.g., to prove (1), use DC in $V$ to construct a countable $S \subset M$ which collapses to $M\_0'$ such that $M\_0' \models \sigma.$ The claim that such a ctm exists is $\Sigma\_2^1$ if I'm not mistaken, so Shoenfield absoluteness implies $M$ also has such a ctm.


So I'm wondering if these claims can be proven directly in ZF; I've heard (2) can be, but I've never seen claim (1) proven without DC. Is there a way to formalize "using choice in an ambient universe" within a model? I know there's a theorem of Woodin that says collapsing a supercompact cardinal in a model of ZF forces DC to hold, but that seems overkill.



Absoluteness, reflection to ctms, and choice in outer models
Yes, all three of these statements can be proved in ZF, without any DC assumption. 


For statement 1, assume $M\models\newcommand\ZF{\text{ZF}}\ZF+\sigma$. By the reflection theorem, there is some ordinal $\theta$ with $(V\_\theta)^M\models\sigma$. One doesn't need DC to prove the reflection theorem, since the argument is about finding an ordinal that is closed under the ranks of witnesses, rather than being able to pick out particular witnesses. 


So in $M$, we have a transitive set, $(V\_\theta)^M$, which is a model of $\sigma$. Let $M[G]$ be a forcing extension of $M$ in which $|V\_\theta|^M$ is countable. In the forcing extension $M[G]$, there is a countable transitive model of $\sigma$, namely, the set $(V\_\theta)^M$, which is countable in $M[G]$. But the assertion "there is a countable transitive model of $\sigma$" is a $\Sigma^1\_2$ assertion, and so it is absolute to $M$. So $M$ has a countable transitive model of $\sigma$, as desired. 


A similar argument works for statement 2. If $M\models \sigma$ and $\sigma$ is $\Sigma\_1$, then by the above, there is a countable transitive model of $\sigma$ in $M$. By Shoenfield absoluteness again, there is a countable transitive model of $\sigma$ in $L$. But if $\sigma$ is $\Sigma\_1$, then $\sigma$ is upward absolute to $L$, and so $L\models\sigma$. 


An essentially similar argument works for statement $3$. Namely, if $M\models\sigma$ and $\sigma$ is $\Sigma\_1$, then there is countable transitive model of $\sigma$, and this countable transitive model is contained in $H\_{\omega\_1}$. So by upward absoluteness of $\Sigma\_1$ assertions, it is true in $H\_{\omega\_1}$, as desired.


