# 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.

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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 $\{0,1,2,\{1\}\}$.

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Nice, thank you! I wonder why I couldn't come up with this. – Rudy the Reindeer Dec 20 '12 at 16:38

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

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I think this is far beyond the scope of the question. If anything the correct example of this kind would be to take a model $M$ of ZF+V$\neq$L, and show that in the inner model $L^M$ there is a set whose power set is different than that of $M$. – Asaf Karagila Dec 20 '12 at 22:02
@AsafKaragila What exactly do you mean by "beyond the scope"? xoxo – Quinn Culver Dec 21 '12 at 0:25
I mean that it's an example of cleaning your house by nuking the entire neighborhood and obliterating any dust particle that was in your house. :-) – Asaf Karagila Dec 21 '12 at 0:27
Dear Quinn, thank you very much. I personally find your example very appropriate and thank you for the reference to Kunen, +1. – Rudy the Reindeer Dec 21 '12 at 7:30

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.)

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Excellent, thank you. I don't understand it, yet, since I haven't reached the Gödel stuff in the book but will soon. – Rudy the Reindeer Dec 21 '12 at 13:40
Matt, I know you haven't. Those questions that you have asked about in the past few days are classical pre-constructibility questions. This is why I pointed out that Quinn's example might be a bit over the top at this moment. – Asaf Karagila Dec 21 '12 at 13:42
I don't know what I love more, your math or your personality! I bet you smell nice too! – Quinn Culver Dec 21 '12 at 18:24
@Quinn: Well I did shower just two hours ago. – Asaf Karagila Dec 21 '12 at 18:27