$U \subset \operatorname{def}(U) \subset P(U)$ for all $U$?

Question

So this is a little bit confusing. I am studying the constructible hierarchy of sets from Jech's set theory. And i think i've found a mistake, please correct me if i'm wrong. After defining $L$ there is this paragraph:

"Note that $U \subset \operatorname{def}(U) \subset P(U)$ for every $U$ so the $L_\alpha$ form a cumulative hierarchy. Thus each $L_\alpha$ is transitive and ..."

So what is confusing me is this that if we assume $U \subset \operatorname{def}(U)$ then we have for all $x \in U \rightarrow x \in \operatorname{def}(U)$ So: $$ x \in U \rightarrow x \in \operatorname{def}(U) = \operatorname{cl}( U \cup \{U\}) \cap P(U) \rightarrow x \in P(U) \rightarrow x \subset U$$This implies that $U$ is transitive. So this is a neccessary condition. But the claim in that paragraph is that this holds for all $U$. And the thing is that the book actually uses this to prove that each $L_\alpha$ is transitive which we clearly need, to prove our premise for $L_\alpha$ to be a cumulative hierarchy.

Thanks for your patience.

Answer 1

This also appears on Jech's third millennium edition, although on page 175, and using $M$ as opposed to $U$.

Indeed, Jech omits the requirement that $M$ is transitive when claiming that $M\subseteq\operatorname{def}(M)\subseteq\mathcal P(M)$. And you are absolutely right, that this requirement is necessary (the inclusion of the definable power set in the power set is obvious, so the first inclusion—$M\subseteq\operatorname{def}(M)$—implies transivitiy). However, since we only really care about $\operatorname{def}(M)$ when $M$ is a transitive set, then this is not a real issue.

It is unfortunate, but when you write a 700+ pages treatise about modern set theory, you are bound to include a few of these type of mistakes. So there's really no reason for alarm.