Prove $\{P(x) :x \in A\}$ is a set

Question

If $A$ is a set how can I prove $\{P(x) :x \in A\}$ is also a set?

I have to use set theory axioms.

($P(X)$ is the powerset of $X$)

Answer 1

One way to do this is by the Axiom of Replacement. The assignment $x\mapsto \mathcal{P}(x)$ is a definable class function, so if $A$ is a set, Replacement tells you directly that $\{\mathcal{P}(x)\mid x\in A\}$ is a set.

But you can also prove this is a set without using Replacement. Note that for all $x\in A$, if $y\in x$, then $y\in \bigcup A$, so $x\subseteq \bigcup A$. Thus if $z\in \mathcal{P}(x)$, then $z\subseteq x\subseteq \bigcup A$, so $z\in \mathcal{P}(\bigcup A)$. Thus $\mathcal{P}(x)\subseteq \mathcal{P}(\bigcup A)$, and hence $\mathcal{P}(x) \in \mathcal{P}(\mathcal{P}(\bigcup A))$. This means we can define $\{\mathcal{P}(x)\mid x\in A\}$ by $$\{B\in \mathcal{P}(\mathcal{P}(\cup A))\mid B = \mathcal{P}(x)\text{ for some }x\in A\}.$$ We used the Axioms of Union, Power Set, and Comprehension.