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.
Best Answer
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 $K \subseteq f[K]$.