Closed set mapped to itself in a compact Hausdorff space

Let $X$ be a compact Hausdorff space and let $f:X\to X$ be continuous. Prove that there exists a non-empty closed set $C$ such that $f(C) = C$.

Proving by contradiction seems hard to me, since if $f(C)\neq C$ for all closed sets $C$, then we can have either $f(C)\subset C$, $f(C)\cap C\neq\varnothing$ or $f(C)\cap C = \varnothing$. The only thing I might have thought about is to set $C = f(X)$. Since $X$ is compact, $f(X)$ is compact and since $X$ is Hausdorff, $f(X)$ is closed. Then we have that $f(C)\subseteq C$ by definition. I was not able to prove that $C\subseteq f(C)$, namely that $f(X) \subseteq f(f(X))$. It is not obvious for me that in general it is even true, so maybe the choice $C = f(X)$ does not help.

• Your guess will not work. Take $X$ to be the closed unit interval $[0,1]$ and $f$ to be the multiplication with $\frac{1}{2}$. – tiefi Jul 27 '17 at 14:07
• Hint: Take the intersection of all nonempty closed subsets $A\subset X$ such that $f(A)\subset A$. – Moishe Kohan Jul 27 '17 at 14:14
• Hint. Define $K_0 = X$ and $K_{n+1} =f(K_n)$. Then $K_0 \supset K_1 \supset K_2 \supset \cdots$ is a sequence of non-empty nested compact sets. What can you say about the intersection $C=\cap_{n\geq 0} K_n$? – Sangchul Lee Jul 27 '17 at 14:23
• @MoisheCohen, How do you know that the intersection is non-empty? For instance, consider $$X=\{x\in\mathbb{R}^2:1\leq|x|\leq2\}, \qquad f(x) =-x.$$ Then you can create two disjoint closed sets $A_1$, $A_2$ for which $f(A_i)\subseteq A_i$ holds for each $i=1,2$. – Sangchul Lee Jul 27 '17 at 14:29
• @tiefi, For instance, you can take $A_1=X\cap(\mathbb{R}\times\{0\})$ and $A_2=X\cap(\{0\}\times\mathbb{R})$ as the intersection of $X$ with the horizontal axis and the vertical axis, respectively. You can check that in fact these sets are invariant under the antipodal map $f$. – Sangchul Lee Jul 27 '17 at 14:37

Indeed define $K_0 = X$. Then define $K_{n+1} = f[K_n]$ for all $n \ge 0$. By induction all $K_n$ are compact subspaces of $X$ and for all $n$: $K_{n+1} \subseteq K_n$:

Both clearly hold for $n=0$. And if $K_n$ is compact and $K_{n+1} \subseteq K_n$, then $K_{n+1} = f[K_n]$ is compact as the continuous image of a compact set, and $K_{n+2} = f[K_{n+1}] \subseteq f[K_n] = K_{n+1}$, as function images preserve inclusions.

Finally define $K=\cap_n K_n$. (it doesn't mattter whether we start from $n=0$ or $n=1$ as $K_0 = X$ has no effect.) Then $K$ is closed as an intersection of closed (compact in Hausdorff implies closed) sets. It is also non-empty as a decreasing intersection of non-empty closed sets in a compact space.

Clearly $$K = \bigcap_{n\ge 1} K_n = \bigcap_{n \ge 1} f[K_{n-1}] \supseteq f[\bigcap_{n \ge 1} K_{n-1}] = f[\bigcap_{n \ge 0} K_n] = f[K]$$

So $f[K] \subseteq K$. To see that the reverse holds:

Let $x \in K$ then for all $n$, $x \in K_{n+1}$ so that $F_n = K_n \cap f^{-1}[\{x\}]$ is non-empty and all $F_n$ are also closed (Hausdorff implies singleton sets are closed) thus compact. So the sets $F_n$ are also a decreasing family of non-empty closed sets and so $$\cap_n F_n \neq \emptyset$$

And note that a $p \in \cap_n F_n$ has the property that $p \in K$ and $f(p) = x$, so that $K \subseteq f[K]$.

• Any reference for the claim "a decreasing intersection of closed sets in compact space has a non-empty intersection"? – Joshhh Jul 30 '17 at 14:17
• @Joshhh non-empty closed subsets. If the intersection were empty, the complements would be an open cover without a finite subcover, contradicting compactness. – Henno Brandsma Jul 30 '17 at 14:51

The answer to your question is given in the comments. I would like to just add the following as an extension to the answers of SangchulLee and MoisheCohen.

Since $X$ is compact, any closed subset $A\subseteq X$ is compact. Since $f$ is continuous, for any compact subset $C\subseteq X$, we have that $f(C)$ is a compact subset of $X$. Finally, as $X$ is Hausdorff, any compact subset of $X$ is closed in $X$.

Now consider the collection $\beta=\{A\subseteq X:A \hspace{2mm}\mbox{closed in} \hspace{2mm} X,\, A\neq\varnothing, \,f(A)\subseteq A\}$. We seek an element $A\in\beta$ such that $f(A)=A$. By the previous paragraph, all elements of $\beta$ are compact and closed subsets of $X$, and for all $B\in\beta$, we have $f(B)\in\beta$.

The idea of MoisheCohen is to take the intersection of all elements of $\beta$. However, this may not give an element of $\beta$, since $\beta$ may contain a pair of disjoint sets. Nevertheless, this idea can be made to work. Suppose we could find $A\in\beta$ such that: if $B\in\beta$ satisfies $B\subseteq A$, then $B=A$. In other words, $A$ is a minimal element of $\beta$. Then by the previous paragraph we have $f(A)\in\beta$ and $f(A)\subseteq A$, whence $f(A)=A$ as required. If we used Zorn's Lemma, we could find such an $A$ and be done. However, we can obtain the desired result by generalising the argument of SangchulLee instead (since here compactness serves as a kind of 'Noetherian' condition).

Claim: If $C\subseteq X$ is closed and satisfies $f(C)\subseteq C$, then we can find $A\subseteq C$ which is also closed in $X$ such that $f(A)=A$.

Proof: We can assume $C\in\beta$ (since for $C=\varnothing$ the result is trivial). Let $C_{1}=C$ and for all $n\geqslant1$ define $C_{n+1}=f(C_{n})$. Observe that $$C_{1}\supseteq C_{2}\supseteq C_{3}\supseteq \cdots$$is a descending chain of non-empty closed sets. Let $A=\cap_{n\geqslant1}C_{n}$.

If $A=\varnothing$, then by taking complements we see that $X=\cup_{n\geqslant1}(X\setminus C_{n})$. By compactness of $X$, we can pass to a finite subcover $X=\cup_{n=1}^{m}(X\setminus C_{n})=X\setminus C_{m}$. But then we have the contradiction that $C_{m}=\varnothing$.

Therefore, we can conclude that $A\neq\varnothing$. Now we just show that $f(A)=A$. It is easy to show from the definitions of the $C_{n}$ that we have $f(A)\subseteq A$.

Now suppose $a\in A$. We seek $b\in A$ such that $f(b)=a$. Note that as $f$ is continuous and $X$ is Hausdorff, $K_{n}:=C_{n}\cap f^{-1}(\{a\})$ is closed in $X$ for all $n$. Each $K_{n}$ is non-empty because $a\in C_{n+1}=f(C_{n})$ for all $n\geqslant1$. Thus, just like the $C_{n}$, the $K_{n}$ form a descending chain of non-empty closed sets. Hence, by a similar compactness argument to when we showed that $A\neq\varnothing$, we see that $K:=\cap_{n\geqslant1}K_{n}\neq\varnothing$. Also $K\subseteq A$, and so for any $b\in K$ we have $a=f(b)\in f(A)$. Therefore $f(A)\supseteq A$. $\hspace{1cm}\square$