How to break power set for non-transitive models? Apparently, $\varphi (x,y) : y = P(x)$ where $P$ denotes the power set is not absolute for transitive models. 
We call a formula $\varphi(v_1, \dots v_n)$ in $L_S$ absolute for a class $\mathbf X$ iff $\forall x_1, \dots , x_n \in \mathbf X (\varphi (x_1 , \dots , x_n) \leftrightarrow \varphi_{/\mathbf X}(x_1, \dots , x_n ))$ where $\varphi_{/\mathbf X}$ denotes the formula $\varphi$ with all occurrences of $\exists v_i$ replaces by $\exists v_i \in \mathbf X$ and similarly for $\forall$.
In words it means that the formula is true regardless of whether we quantify over the whole universe $\mathbf V$ or only over elements of $\mathbf X$.
It is easy to break $\varphi (x,y) : y = P(x)$ in non-transitive models: for example if $M = \{ a, \{\varnothing , a\}, \{\varnothing , a, b\} \}$ then in $M$, $\{\varnothing , a, b\}$ is a power set of $a$ but outside $M$ this is false. 
Can someone show me a transitive model in which $\varphi (x,y) : y = P(x)$ is not absolute? Many thanks for your help.
 A: It's not a model of very many of the ZF axioms, but consider the "model" consisting of the sets


*

*$0=\{\}$

*$1=\{0\}$

*$2=\{0,1\}$

*$3=\{0,1,2\}$


and nothing else. Relative to this model it is true that $3=\mathcal P(2)$, because the members of $3$ are exactly the objects of the model that are subsets of $2$.
However, $3=\mathcal P(2)$ is not true outside the model, because in reality $\mathcal P(2)$ is $\{0,1,2,\{1\}\}$.
A: Another, more (set-theoretically) natural example comes from exercise (B8) in chapter VII of Kunen's Set Theory: An Introduction to Indpedence Proofs. There one proves that if you start with a countable transitive model of ZFC, and force with a non-trivial (in this case, non-atomic) partial order countably many times, then the union of all the forcing extensions cannot be a model of ZFC, precisely because it fails to satisfy the power set axiom (even though each extension does). The reason is that even though each generic filter is in the union, the set containing all of them isn't, so the second power set of the partial order doesn't exists.
Love,
Quinn
A: Since forcing was deemed a reasonable example, here is the "opposite" example.
Suppose that $(M,\in)$ is a transitive model of ZFC and $M\models V\neq L$, that is $M$ is not a model of Godel's axiom of constructibility. Let $L^M\subseteq M$ be the inner model of $M$ which satisfies the axiom of constructibility, then $(L^M,\in)$ is also a transitive model of ZFC, and it is a substructure of $(M,\in)$.
It is a nontrivial claim, but one can show that there exists a set of ordinals $A$ such that $A\in M\setminus L^M$, that is a set of ordinals which is non-constructible in $M$. 
Let $\delta=\sup A$ then $\mathcal P^M(\delta)\neq\mathcal P^{L^M}(\delta)$. This follows trivially from the above, as $A\subseteq\delta$ is in $M$ but not in $L^M$.

In fact the above can be generalized and we can show that every whenever $N\subseteq M$ is an inner model of $M$ either $M=N$ or there is an ordinal $\delta$ such that $\mathcal P^M(\delta)\neq\mathcal P^N(\delta)$.
This follows from the following theorem,

Theorem. Let $(M,\in)$ and $(N,\in)$ be two [transitive] models of ZFC with the same ordinals and the same sets of ordinals then $M=N$.

(I will add a minor remark that the assumption of ZFC is needed, it is consistent that there are two models of ZF with the same ordinals and the same sets of ordinals which are different from each other.)
