let $A$ be any inductive set, then $\{C \in P(A)|C \text{ is inductive set} \}$ is a set? ... and $\mathbb{N}$...?

Question

let $A$ be any inductive set, then $\{C \in P(A)|C \text{ is inductive set} \}$ is a set?

if $\{C \in P(A)|C \text{ is inductive set} \}$ is a set I can defined $\mathbb{N}:=\bigcap\{C \in P(A)|C \text{ is inductive set} \}$??

Thanks in advance!!

P.S.=$P(A)$ is power of (A)

Answer 1

Yes, $\mathscr{I}=\{C\in\wp(A):C\text{ is an inductive set}\}$ is a set: $\wp(A)$ is a set by the power set axiom, and $\mathscr{I}$ is then a set by the axiom schema of comprehension (or separation, or specification). (Of course this requires verifying that ‘$x\text{ is an inductive set}$’ can be expressed as a formula in the language of set theory, but that’s straightforward.)

You can then form $M=\bigcap\mathscr{I}$. (Since $A\in\mathscr{I}$, it’s clear that $M\subseteq A$.) To finish the argument, you’ll need to show that $M$ is inductive, and that if $I$ is any inductive set, then $M\subseteq I$. The first of these is completely straightforward. For the second, show that $I\cap A$ is inductive.

Answer 2

Yes. We can write a formula $\varphi(x)$ which states that $x$ is an inductive set. The power set axiom assures us that $\mathcal P(A)$ is a set, and so the separation schema assures us that the inductive subsets of $A$ is also a set.

Finally, note that $\bigcap\{\ldots\}$ is not intrinsic to the axioms of $\sf ZFC$ (rather the union is), so you just need to assure that the set you are intersecting is not empty. However since $A$ is inductive, it is not empty, and we're good.