Given a transitive model $M$, is it possible to have both $X \in M$ and $X^M \not = X$?

Question

Suppose $M$ is a transitive model (set or proper class) of, say, $ZFC$ (weaken or strengthen the theory if you wish). Is it possible to have a set $X$ such that $X \in M$ but also $X^M \not = X$? Here $X^M = \{x \in M: \varphi^M(x)\}$ where $\varphi$ is the defining formula for $X$.

I think not but I can't prove it. I can't think of an example of such an $X$ either.

Answer 1

Yes, this can certainly happen. Here's a very easy example: consider the set $X$ defined by $X=0$ if $2^{\aleph_0}=\aleph_1$ and $X=1$ otherwise. Then $X$ is certainly in any transitive model $M$ of ZFC, since both $0$ and $1$ are. But $X^M$ need not be the same as $X$, since $M$ need not agree with $V$ about CH.

Answer 2

Yes. Consider cardinals: $\aleph_1\in L$ but frequently $\aleph_1^{L}\neq \aleph_1$. (In particular, if we force with the poset collapsing the ordinal $\aleph_1^L$ to be countable, the generic extension $V[G]$ satisfies $\aleph_1^L\neq \aleph_1^{V[G]}$.)