$f\left( \bigcap_{n=1}^{\infty}A_{n}\right) = \bigcap_{n=1}^{\infty}f(A_{n})$ for a continuous function on a compact space

Question

I have the following statement.

Let $X$ a compact metric space. Let $f: X \rightarrow X$ a continuous function. If $\lbrace A_{n} \rbrace$ is a decreasing sequence of nonempty closed sets on $X$, then

$$f\left( \bigcap_{n=1}^{\infty}A_{n}\right) = \bigcap_{n=1}^{\infty}f(A_{n}) $$

Now, it is easy to see that

$$f\left( \bigcap_{n=1}^{\infty}A_{n}\right) \subseteq \bigcap_{n=1}^{\infty}f(A_{n}) $$

Could someone give a suggestion to get the other part?

Answer 1

Hint Let $y \in \bigcap_{n=1}^{\infty}f(A_{n})$. Then, there exists some $x_n \in A_n$ such that $f(x_n)=y$.

Now, by the compactness of $X$, $x_n$ has a convergent subsequence $x_{k_n}\to x$. By the continuity of $f$ you have $f(x)=y$.

Show that $x \in \bigcap_n A_n$.

Answer 2

As a follow-up to the accepted answer: Let $X$ be $T_1$ and compact. Let $f:X\to X$ be continuous. Let $\Bbb A$ be a family of closed subsets of $X,$ where $\Bbb A$ has the Finite Intersection Property (FIP). That is, if $\Bbb B$ is a non-empty finite subset of $\Bbb A$ then $\cap \Bbb B\ne \emptyset.$

Suppose $y\in \cap_{A\in \Bbb A}f(A).$ Then $y\in f(\cap_{A\in \Bbb A}A).$

Proof: For $A\in \Bbb A$ let $A^*=A\cap f^{-1}\{y\}.$ Now $\{y\}$ is closed (because $X$ is $T_1)$ so $f^{-1}\{y\}$ is closed (because $f$ is continuous). Hence $F^*=\{A^*:A\in \Bbb A\}$ is a non-empty family of closed subsets of $X$ and it is easily verified that $F^*$ has the FIP.

Exercise: A space $X$ is compact iff every non-empty family $\Bbb H$ of closed sets, such that $\Bbb H$ has the FIP, satisfies $\cap H\ne \emptyset.$

So there exists $x\in \cap F^*$ because $X$ is compact. That is, $\forall A\in \Bbb A\;(f(x)=y\land x\in A\}.$ That is, $x\in \cap \Bbb A$ and $f(x)=y,$ so $y\in f(\cap_{A\in \Bbb A} A).$

APPENDIX. (Proof of "Exercise".) Remark: Every statement about open sets has an equivalent "dual" statement about their complements, the closed sets. The property in the Exercise can be considered the dual of "Every open cover of $X$ has a finite subcover". In the Exercise: (i) Let $\Bbb J$ be the family of non-empty finite subsets of $\Bbb H.$ If $\cap \Bbb H=\emptyset$ then $\{X\setminus \cap J: J\in \Bbb J\}$ is an open cover of $X$ with no finite subcover, so $X$ is not compact. (ii). Conversely if $X$ is not compact, let $\Bbb C$ be an open cover of $X$ with no finite sub-cover. Let $\Bbb D$ be the family of non-empty finite subsets of $\Bbb C.$ Then $\Bbb I=\{X\setminus \cup D: D\in \Bbb D\}$ is a family of closed sets such that $\Bbb I$ has the FIP, but $\cap \Bbb I=\emptyset .$