This is exercise 13.18 of Jech's Set Theory:

If $M \prec L_{\omega_2}$, then $\omega_1 \cap M = \alpha$ for some $\alpha \leq \omega_1$.

The previous exercise is very similar and has been discussed on math.SE already: If $M\prec L_{\omega_1}$ then $M = L_\alpha$ for some $\alpha$ - we need a condition to prove it?

Reading the question linked above, there seems to be a mistake in the statement. So I wonder: is the same mistake present here? Would a correct version be as follows?

If $M \prec L_{\omega_2^L}$ with ($M$ countable)$^L$, then $\omega_1^L \cap M = \alpha$ for some $\alpha < \omega_1^L$.

Also, how can I prove it? If $\gamma < \omega_1^L$ then there is a surjection $f \colon \omega \to \gamma$ which belongs to $L$. If I can show that any such $f$ belongs to $L_{\omega_2}$, then I know how to conclude the proof. But how can I be sure that at least one surjection $\omega \to \gamma$ is in $L_{\omega_2}$?

  • $\begingroup$ I suppose that you (Jech) mean(s) that $M$ is countable. $\endgroup$ – Asaf Karagila Aug 26 '16 at 19:09
  • $\begingroup$ Sure, my mistake. Fixed. Thanks $\endgroup$ – aerdna91 Aug 26 '16 at 21:53

Yes, this should be $\omega_1^L$ instead of $\omega_1$, etc. To see why this is necessary, suppose $\omega_1^V\ge\omega_2^L$. Now take a countable in $L$ elementary submodel $M$ of $L_{\omega_2}$ with $\omega_1^L\in M$. (We can do this since $\omega_1^L, L_{\omega_2}$ are each in $L$, and $L$ verifies the Lowenheim-Skolem theorem.) This $M$ must not contain all of $\omega_1^L$, since $M$ is countable in $L$, and so $\omega_1\cap M$ is not closed downwards.

As to the proof, since $\gamma$ is a ($L$-)countable ordinal, the function $f$ is ($L$-)hereditarily countable. By condensation, then, $f\in L_{\omega_1^L}$. This is for exactly the same reason that $\mathbb{R}\cap L\subset L_{\omega_1}$. (I don't see why $\omega_2^L$ is necessary, in fact.)

  • $\begingroup$ Great, thanks. Could you please elaborate on the use of condensation in your proof? My strategy would be to use the fact that $V = L \Rightarrow L_{\omega_1} = H_{\omega_1}$, so $(f \in H_{\omega_1})^L \Rightarrow (f \in L_{\omega_1})^L$. But this is probably overkill and I would like to understand how it works at a deeper level. $\endgroup$ – aerdna91 Aug 26 '16 at 14:26
  • $\begingroup$ @aerdna91 That's all I was thinking. $\endgroup$ – Noah Schweber Aug 26 '16 at 14:31

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