Monotonicity Question in Enderton's text on Set Theory I am struggling to begin solving the following question in Enderton's textbook on set-theory:

Assume that $F: P(A) \rightarrow P(A)$ and that $F$ has the monotonicity property: 
$X \subseteq Y \subseteq A \ \implies \ F(X) \subseteq F(Y)$.
Define: $B = \bigcap \{ X \subseteq A \ | \ F(X) \subseteq X\}$ and $C = \bigcup  \{ X \subseteq A \ | \ X \subseteq F(X)\}$.
(a) Show that $F(B) = B$ and $F(C) = C$.
(b) Show that if $F(X) = X$, then $B \subseteq X \subseteq C$.

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Something I have tried so far is to assume that $x \in F(B)$ in order to show that $x \in B$. Beginning with this, I get $x\in$ ran $F$, and so $x = X$ for some $X \subseteq A$.  I also have, $F(B) \subseteq$ ran $F$ however I'm unsure how to proceed from here.
Any advice anyone can provide would be much appreciated.
Thank you in advance.
 A: Let us denote
$$\mathcal B=\{X\subseteq A; F(X)\subseteq X\}.$$
Then $B$ is simply the intersection of this system, i.e., $B=\bigcap \mathcal B$.
Notice that $\mathcal B$ is non-empty, since $A\in\mathcal B$.

(a) Notice that for each $X\in\mathcal B$ we have $B\subseteq X$. By monotonicity this implies $F(B)\subseteq F(X)$.
Together we get
$$(\forall X\in\mathcal B)F(B)\subseteq F(X)\subseteq X$$
which implies
$$F(B)\subseteq \bigcap \mathcal B=B.$$
(Since $F(B)$ is subset of every set in this system, it is also subset of the intersection.)
As we have already shown that $F(B)\subseteq B$, then monotonicity implies $F(F(B))\subseteq F(B)$. But this means that
$$F(B)\in\mathcal B,$$
hence
$$B=\bigcap\mathcal B \subseteq F(B).$$
(Intersection is subset of every set in the system.)

You can show the second part of (a) similarly; or simply just notice that it is the dual claim to the claim we have already shown.

(b) If $F(X)=X$, then $X\in\mathcal B$, therefore
$$B=\bigcap\mathcal B\subseteq X.$$
A: (a): one side:let $x\in F(B)$.  then there is $b\in B,F(b)=x$.so for each$X\subseteq A,F(X)\subseteq X,b\in X$.
so for each $X$ from those$ F(b)\in F(X)$ so $x=F(b)\in \cap\{X\subseteq A|F(x)\subseteq X\}=B$.so $F(B)\subseteq B$.
second side:let $b\in B$. then for each $X\subseteq A $ that $F(X)\subseteq X,b\in X$
,so for each of those $X,F(b)\in F(X)\subseteq X$ ,so $F(b)\in B$,and then $F(B)\subseteq B$ 
so $F(B)=B$
(b)one side:
 let $c\in C$. for each $X\subseteq A,X\subseteq F(X)$ gives $X\subseteq C$.
so there is a set $X\subseteq A,X\subseteq F(X)$ such that $c\in X$. and then$ c\in X\subseteq F(X)\subseteq F(C)$, so$ C\subseteq F(C)$.
and the other:lemma: if $X\subseteq C$ then $F(X)\subseteq C$.
**proof:**if $X\subseteq C$ then $X\subseteq F(X)$,so $F(X)\subseteq F(F(X))$.
now, let $x\in F(c)$. there is a set $X\subseteq A,X\subseteq F(X)$ that there is $c\in X,F(c)=x$. so $x\in F(X)$ and by the lemma $x\in C$. so $F(C)\subseteq C$
so $F(C)=C$.
