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.)
Best Answer
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:
As you point out, we may assume $M=L_\alpha$.
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.
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.