# Continuity of Monotonic Set Functions

Consider a function $$F: 2^X \to 2^X$$, where $$2^X$$ is the powerset of $$X$$. The function $$F$$ is monotone, if it preserves inclusion. That is, if $$A \subseteq B \subseteq X$$, then $$F(A) \subseteq F(B) \subseteq X$$.

My question is, if I have a decreasing sequence $$P_0 \supseteq P_1 \supseteq \ldots$$, is it always true that $$\bigcap_{k=0}^\infty F(P_k) = F\left(\bigcap_{k=0}^\infty P_k\right).$$

If it is not always true, then under what additional assumption would it be true?

Thanks a lot for the help!

I can show one direction of the inclusion. $$F\left(\bigcap_{k=0}^\infty P_k\right) \subseteq F\left(\bigcap_{k=0}^K P_k\right) = F(P_K) \quad \forall K\geq 0.$$ Then take the infinite intersection on the right hand side, I have $$F\left(\bigcap_{k=0}^\infty P_k\right) \subseteq \bigcap_{k=0}^\infty F(P_k).$$

Also, it was quite attempting to do a proof by induction, since we have $$\bigcap_{k=0}^K F(P_k) = F\left(\bigcap_{k=0}^KP_k\right) \quad \forall~ K\geq 0.$$ But, I am aware that proof by induction only holds for arbitrary large finite integer...

This is not true. Take $$X=\mathbb N.$$ Chose an arbitrary $$Z\neq \emptyset\in 2^\mathbb N.$$ Define F as follows: for all $$Y\neq \emptyset \in 2^\mathbb N$$ $$F(\emptyset)=\emptyset,F(Y)=Z$$ Then F is weakly monotone since $$\emptyset \subseteq Z\subseteq X$$. Now take $$P_k=\{x\in \mathbb N |x\ge k\}.$$ The $$P_k$$ are decreasing, but $$\bigcap_{k=0}^\infty F(P_k)=\bigcap_{k=0}^\infty Z = Z\neq \emptyset =F(\emptyset)=F(\bigcap_{k=0}^\infty P_k).$$ This might hold if you assume that $$\bigcap_{k=0}^\infty P_k$$ is non-empty, but I am not sure.
I'd like to provide another example. Let $$X$$ be a topological space. Define $$F_1(A)=\overline A$$ and $$F_2(A)=\mathrm{int}\, A$$ for all $$A\subset X$$. Then closure $$F_1$$ doesn't behave well for increasing sequence and $$F_2$$ with decreasing, although both of them are monotone.
For example, let $$r_k=1+1/k$$ and $$P_k= [0,r_k]$$. Then $$P_1\supset P_2\supset\ldots$$. Moreover, $$F_2(P_k) = (0,r_k),\quad \bigcap_{k=1}^\infty F_2\left( P_k\right)= (0,1],\quad F_2\left(\bigcap_{k=1}^\infty P_k\right) = F_2([0,1])=(0,1).$$