Let $X$ be compact Hausdorff space and $f:X \to X$ a continuous map. Show that there exists a non-empty closed subset $A$ of $X$ such that $f(A)=A.$

Let $X$ be non-empty compact Hausdorff space and $f:X \to X$ a continuous map. Show that there exists a non-empty closed subset $A$ of $X$ such that $f(A)=A.$

If $f$ is continuous then it maps compact sets to compact sets. Also since $X$ is compact any closed subset of $X$ is compact implying that $A$ is compact.

I feel like I’ve missed the part of writing proofs that assesses how to prove the existence of some object? Should I consider a contradiction and suppose such $A$ doesn’t exists? I don’t think I have anything to work directly from $f(A)=A$?

Hint:

• Let $A_0=X$ and recursively $A_{n+1}=f(A_n)$.
• Show that $A_0\supseteq A_1\supseteq A_2\supseteq\ldots$
• Let $A=\bigcap_nA_n$.
• Show $f(A)=A$.
• Show that $A$ is a non-empty closed subset.