Axiom of Replacement in the cumulative hierarchy of sets

Question

I am reading the proof for the consistency of the axiom of Regularity relative to ZF - Regularity and what the book does is consider the cumulative hierarchy: $$P = \bigcup\limits_{\alpha \in Ord} V_\alpha$$ And prove that $P \models Extensionality, P \models Parinig, ...$ but the thing that confuses me is when we reach Replacement. So we have to prove: $$ P \models \forall x \forall y \forall z [\varphi(x, y, p) \wedge \varphi(x, z, p) \rightarrow y = z] \rightarrow \forall X \exists Y \forall y [ y \in Y \leftrightarrow (\exists x \in X) \varphi(x, y, p)]$$ Given a $p \in P$ and assuming the first part holds, we get a function, $F = \{ (x, y) \in P: P \models \varphi(x, y, p)\}$. Now we let $Y = F[X]$. (This part is confusing me): Since $Y \subset P$, we have $Y \in P$.

Why? Why cant $Y$ be a proper subclass of $P$? How is this implication justified?

Thanks for your patience.

Answer 1

You need to use the fact that Replacement holds in the full universe you're working in.

Now look at a formula whose relativization to $P$ is functional, and apply Replacement outside of $P$. This gives you a set, so all you need to do is show that it lies inside $P$. But because you have a set, the ranks of its members only form a set of ordinals, so it must be a set in $P$.