Continuous map from a hausdorff space to itself Let X be a compact Hausdorff space, and let  $f:X \to X$ be a continuous map of X into itself.  Prove that there is a non-empty subset A of X such that $f(A)=A$.
If I let $A_1 = f(X)$ and inductively $A_{n+1}=f(A_n)$. Let $A=\cap _{n=1}^\infty A_n $
Why does this work?
 A: Prove by induction that $A_{n+1} \subset A_n$. Let $y \in A$. Then for each $n$, $y \in A_{n+1}$ so $y=f(x_n)$ with $x_n \in A_n$. By compactness there is a subnet of $\{x_n\}$ which converges to some point $x$. Note that $x_{n+m} \in A_{n+m} \subset A_n$ so $x \in A_n$ for each $n$. [ Of course the sets $A_n$ are all compact, hence closed]. By continuity of $f$ we also have $y=f(x)$. Hence $y \in f(A)$. We have proved that $A \subset f(A)$. The reverse inclusion is easy. If more details are needed let me know.
A: Nitpicking 101: The empty space is compact Hausdorff. So assume $X\ne \emptyset.$
Let $X_0=X$ and $X_{n+1}=f(X_n)$ for $n\geq 0.$ Obviously $A_1\subset A_0.$ If $A_{n+1}\subset A_n$ then $$A_{n+2}=f(A_{n+1})=f(A_{n+1}\cap A_n)\subset f(A_n)=A_{n+1}.$$ So by induction, $A_{n+1}\subset A_n$ for all $n\geq 0.$
Any statement about open sets has an equivalent "dual" statement about closed sets. An equivalent to "Every open cover of $X$ has a finite sub-cover" is:
$(\bullet).$ If $F$ is a non-empty closed family in $X$ and $F$ has the FIP then $\cap F\ne \emptyset.$ 
Translation: A closed family in $X$ is a family of closed subsets of $X...$  FIP is Finite Intersection Property, which means that if $G$ is a non-empty finite subset of $F$ then $\cap G\ne \emptyset.$
Let $F=\{A_n:n\geq 0\}$. Assuming $X\ne \emptyset,$ we have (fairly obviously) $A_n\ne \emptyset\implies A_{n+1}= f(A_n)\ne \emptyset,$  so $F$ has the FIP. Now the continuous image of a compact space is compact, so by induction on $n\geq 0,$ every $A_n$ is compact. And since $X$ is Hausdorff, compact sub-spaces of $X$ are closed in $X$. Hence by $(\bullet )$ we have $A=\cap F\ne \emptyset.$
It should be fairly obvious that $f(A)\subset A.$ To show that $A\subset f(A),$ take any  $a\in A,$ and  for $n\geq 0$  let $H_{n,a}=A_n\cap f^{-1}\{a\}.$ We have  $$a\in A_{n+1}=f(A_n)\implies \emptyset \ne A_n \cap f^{-1}\{a\}=H_{n,a}.$$ Now $H_a=\{H_{n,a}: n\geq 0\}$ is a non-empty closed family in $X$ with the FIP so by $(\bullet )$ we have $\cap H_a \ne \emptyset.$  So take any  $b\in \cap H_a.$ Then $f(b)=a$ and $b\in \cap_{n\geq 0}A_n=A,$ so $a\in f(A).$ 
Remark: $\{a\}$ is closed  because $X$ is Hausdorff. And $f$ is continuous so $f^{-1}C$ is closed when $C$ is closed . So $f^{-1}\{a\}$ is closed. And every $A_n$ is closed. So $H_{n,a}$ is closed.
