# Ways to construct set models of $\mathsf{ZFC}$ from existing ones

So I'm trying to solve this exercise, but I am having trouble solving it without any further assumptions. This is the exercise:

Assume $$M$$ is a transitive model of $$\mathsf{ZFC}$$ such that if $$\alpha = M\cap \mbox{Ord}$$, then $$|\alpha| = \aleph_1$$. Show that there exists a transitive model $$M'$$ of $$\mathsf{ZFC}$$ such that $$M'\cap \mbox{Ord} = \alpha$$ and $$M'\neq M$$.

Now if $$M\neq L_\alpha$$, we can take $$M' = L_\alpha$$. Otherwise if $$M = L_\alpha$$ and $$(\omega_1)^M < \omega_1$$, using the Rasiowa-Sikorski lemma we can force over $$M$$ with the Cohen forcing and take $$M' = M[G]$$. But if $$(\omega_1)^M = \omega_1$$, I don't see a clear way of using forcing since we don't have $$\mathsf{MA} + 2^{\aleph_0} > \omega_1$$ or any other helpful hypothesis. I'm stuck here.

One other semi-idea I had, was to take some $$X\in V_\alpha - M$$ and consider some sort of constructible closure of $$M \cup \{X\}$$, but that didn't work out either.

What are some other ways of constructing models of set theory without much assumptions? (Aside from Lowenheim-Skolem theroem which may give ill-founded models.)

• Where did you see this question, by the way? Dec 2, 2020 at 17:46
• @AsafKaragila, It is an exercise in Ralf Schindler's book. To be precise, it's question $6.11$ from the $6$th chapter. Dec 2, 2020 at 17:48
• Funny. I guess it's a natural question. Dec 2, 2020 at 17:58

This was a question I asked many years ago on MathOverflow. Namely, can there be only one transitive model of an uncountable height.

Françios Dorais gave a fantastic answer:

1. As you point out, we may assume $$M=L_\alpha$$.

2. First option, $$\alpha=\omega_1$$, in that case every $$x\in M$$ is countable in $$V$$, so the Cohen forcing over $$M$$ has only countably many dense open sets. Therefore we can find a generic over it in the usual sense.

3. Well, the first option failed. So $$\omega_1\in M$$. In that case, $$\operatorname{Add}(\omega_1,1)^L\in M$$, and it is $$\sigma$$-closed, and since $$|M|=|L_\alpha|=|\alpha|=|\aleph_1|$$, there are only $$\aleph_1$$ dense open sets in $$M$$ itself. So we can diagonalise over them to produce a generic.

This is the same usual Rasiowa–Sikorski lemma, where at limit steps you use the $$\sigma$$-closure.

• +1 Beat me to it! Dec 2, 2020 at 17:39
• Well, François beat both of us to it. :) Dec 2, 2020 at 17:40
• Now that I am comparing our questions, mine seems like a word for word copy of yours! :) I wonder why it didn't come up in my searches. Thanks a lot! Dec 2, 2020 at 17:47