# An uncountable transitive model believes all reals are constructible

I'm struggling to solve the following old exam problem:

Assume $\mathsf{ZFC}+V=L$. Prove that if $M$ is any uncountable transitive set such that $(M,\in)\models\mathsf{ZFC}$, then $(M,\in)\models$ "every set of integers is constructible".

I think it should be a standard Condensation argument, but I can't get it to work. A couple of observations:

1. It is a standard Condensation argument easy if $M$ (rather than $V$) satisfies $V=L$.
2. The proof that the $\mathsf{GCH}$ holds in $L$ gives that every constructible subset of $L_{\omega}$ is a member of $L_{\omega_1}$. Assuming $V=L$, this means that $P(\omega)\subseteq L_{\omega_1}$. Therefore it suffices to show (though I don't see how, and I'm not sure that it's necessarily true) that every countable ordinal belongs to $M$, so that $M$ contains $L_{\omega_1}$. If this were true, then $M$ would not only believe that every real is constructible; it would contain every constructible real and the ordinal witnessing its constructibility.
3. $L^M = L_\alpha$ for some $\alpha$.
4. If $\sup(\mathrm{ON}\cap M)$ is countable, then (since $M$ is uncountable) $M$ must contain sets of arbitrarily large $L$-rank below $\omega_1$. Again, it seems unlikely that $\sup(\mathrm{ON}\cap M)$ could be countable, but I don't see how to prove that it can't be.

Anyone see what I'm missing?

• By "a couple of" I meant "four". Feb 22 '13 at 7:08
• If $M$ satisfies $V=L$, then $M$ satisfies "every set (of integers or other things) is constructible" by definition, not by a condensation argument. Feb 22 '13 at 7:10
• You're right, of course. Good point. Feb 22 '13 at 7:11

If the height of $M$ is countable, then some $V_\alpha^M$ is uncountable, but then it cannot be in bijection with an ordinal of $M$. This is a contradiction.
It follows that the height of $M$ is uncountable, but then you are done, because $L^M$ is $L_\alpha$ for some uncountable $\alpha$. But $L_{\omega_1}$ already contains all reals.