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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...

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2 Answers 2

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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.

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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, $$F_2(P_k) = (0,r_k),\quad \bigcap_{k=1}^\infty F_2\left( P_k\right)= (0,1],\quad F_2\left(\bigcap_{k=1}^\infty P_k\right) = F_2([0,1])=(0,1).$$

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