To make the title a little bit more proecise:
Let $X$ be a set and a map, $F:\mathcal{P} (X) \rightarrow \mathcal{P} (X)$ (where $\mathcal{P} (X)$ denotes the powerset of $X$), such that $A \subseteq F(A)$ and $F(A) \subseteq F(B)$ for all $\mathcal{P} (X) \ni A \subseteq B \in \mathcal{P} (X)$.
If we call a set $T \in \mathcal{P} (X)$ closed if $F(T)=T$.
How can I prove that given a function $F$ like above, a map F':\mathcal{P} (X) \rightarrow \mathcal{P} (X) exists, such that all sets that are closed under $F$ are also closed under F' and such that F' satisfies the same properties as $F$ and additionally the property F'(F'(A))=F'(A) for all $A \in \mathcal{P} (X)$ ?
Somehow $F$ would have to be iterated infinitely many times, to obtain the idempotency, but I can't figure out, how to do that.