# Extension of ZFC models preserves cardinals

Let $M \subseteq N$ be countable transitive ZFC set models. Assume that this extension preserves cardinals, i.e. if $\alpha$ is an ordinal number (this notion is absolute) such that $(\alpha \text{ is a cardinal})^M$, then also $(\alpha \text{ is a cardinal})^N$. Notice that the other direction holds in any case.

Now I think that for every ordinal $\alpha$ we have $\aleph_\alpha^M = \aleph_\alpha^N$. The proof uses induction by $\alpha$. The case $\alpha=0$ follows since $\omega$ is absolute. The successor step uses the assumption. The limit step uses that $\sup$ is absolute. Can someone confirm this sketch of proof?

Background: In Kunen's Set theory this seems to be used in the context of Cohen forcing. I wonder why it isn't mentioned explicitly.

• I think you are assuming that the set of ordinals of $M$ and of $N$ are the same, and in the second paragraph you mean for every ordinal $\alpha \in M.$ This of course is the case when $N$ is a forcing extension of $M.$ I suspect Kunen doesn't mention it because he thought it obvious, although it might be mentioned in one of the exercises in Chapter 7. – DanielWainfleet Jun 16 '16 at 4:10

## 1 Answer

Yes, you can see this in a slightly more explicit way:

Consider $\kappa$ which is a cardinal in $M$ and $N$, then $A_M=\{\beta<\kappa\mid M\models\omega\leq\beta,\beta\text{ is a cardinal}\}$, and $A_N$ defined similarly, are the same by the induction hypothesis. If the two sets are the same, their order type is the same, and it is clear that $\kappa=(\aleph_{\operatorname{otp}(A_X)})^X$ for $X\in\{M,N\}$.