Function defined on power set Let $E$ be a set and let $f:\mathcal P(E)  \to \mathcal P(E)$ such that if $A \subset B \implies f(A) \subset f(B)$. We define $R=\{Z \subset E :f(Z)\subset Z\}$ and $S=\{Z \subset E : Z \subset f(Z)\}$. We consider the subsets  $$V=\bigcap_{Z \in R}Z, \space W=\bigcup_{Z \in S} Z$$
Show that $f(V)=V$ and $f(W)=W$.
I got stuck trying to show one inclusion in both cases, I'll write what I did up to now:
$$f(V)=f(\bigcap_{Z \in R}Z) \subset \bigcap_{Z \in R}f(Z) \subset \bigcap_{Z \in R}Z=V,$$$$W=\bigcup_{Z \in S} Z \subset \bigcup_{Z \in S} f(Z)=f(\bigcup_{Z \in S} Z)=f(W)$$
I don't know what to do to show the other inclusions, I would appreciate any small hints rather than a complete answer. Thanks in advance.

Since you've shown that $f(V)\subseteq V$, monotonicity gives you $f(f(V))\subseteq f(V)$.  So you have $f(V)\in R$ and therefore $V=\bigcap_{Z\in R}Z\subseteq f(V)$.
