Function on a Power Set Let $f\colon \mathcal{P}(A)\mapsto \mathcal{P}(A)$ be a function such that $U \subseteq V$ implies $f(U) \subseteq f(V)$ for every $U, V \in \mathcal{P}(A)$. Show there exists a $W \in \mathcal{P}(A)$ such that $f(W) = W$.
This is what I've been thinking:
Notice $A \subseteq A$ therefore $f(A) \subseteq f(A)$ and as $f(A) \in  \mathcal{P}(A)$, this implies $f(A) \subseteq A$.
Then $f(f(A)) \subseteq f(A) \subseteq A$ and so $f(f(f(A))) \subseteq f(f(A)) \subseteq f(A) \subseteq A$.
If $A$ is finite, this process should leave you with the desired $W$ (I think) after a finite number of iterations. Not so sure about the infinite case.
I might even be going about this totally wrong so any suggestions would be very much appreciated. Thanks!

This is the Knaster-Tarski theorem, actually. Let me give you a hint forward.
And you're essentially on the right track, but instead of constructing it transfinitely, what happens when you look at all the sets $\{B\mid f(B)\subseteq B\}$? What would their intersection be?
