Let $F:A \to A$ be a function and let $Y \subseteq A$. Prove that the class $S = \{B:Y\subseteq B \subseteq A$ and $F[B] \subseteq B\}$ is a set.

Question

Let $F:A \to A$ be a function and let $Y \subseteq A$.

(a) Prove that the class $S = \{B:Y\subseteq B \subseteq A$ and $F[B] \subseteq B\}$ is a set.

Can I use the theorem that states: Let $\phi(x)$ be a formula. Suppose that there is a set $A$ such that for all $x$, if $\phi(x)$, then $x \in A$. Then there is a unique set $D$ such that for all $x$, $x \in D$ iff $\phi(x)$.

Answer 1

Here's a hint: replace the letter $A$ in your theorem with $U$, so that it reads

Let $\phi(x)$ be a formula. Suppose that there is a set $U$ such that for all $x$, if $\phi(x)$, then $x \in U$. Then there is a unique set $D$ such that for all $x$, $x \in D$ iff $\phi(x)$.

Rephrasing, that says that if you have some condition $\phi$ with the property that anything for which $\phi(x)$ is true is also in the set $U$ (but $U$ might have lots of other stuff in it), then you can also form the subset of $U$ consisting of things $x$ for which $\phi(x)$ is true.

The set you're trying to construct (the class named $S$) consists of a bunch of items generically named $B$. What kind of item is such a $B$?