# Compact Hausdorff space with continuous function

1. Oct 15, 2005

### Oxymoron

Question
Let $X$ be a compact Hausdorff space and let $f:X\rightarrow X$ be continuous. Show that there exists a non-empty subset $A \subseteq X$ such that $f(A) = A$.
At the moment I am trying to show that $f$ is a homeomorphism and maybe after that I can show that $f(A) = A$. But Im not sure if this is the right tactic.
I know that $f$ is a continuous function, and since $f:X \rightarrow X$ it is bijective (?) I thought it might be since the domain and range coincide.
Anyway, if I take a closed subset $A \subseteq X$ then $A$ is automatically compact (since every closed subset of a Hausdorff space is compact). Then since $f$ is continuous it maps closed compact sets to closed compact sets.
Since $f$ is bijective, its inverse $f^{-1}$ exists, and $f^{-1}$ will also map closed compact sets to closed compact sets.
Therefore $f$ is a homeomorphism. Im not sure if I can conclude from this that $A$ is homeomorphic to $f(A)$ which implies $f(A) = A$.
PS. What about the open sets in $X$? I need help!

Last edited: Oct 15, 2005
2. Oct 15, 2005

### fourier jr

just because a continuous map goes from a set to itself doesn't mean it's a bijection. consider f:X->X with f(x)=0 for every x in X; that's continuous but not a bijection. off the top of my head i would try using the fact that a) every closed subspace of a Hausdorff space is compact and b) that every compact subspace of a Hausdorff space is closed. maybe also use the fact that a continuous image of a compact space is compact. i'm not sure if that will help or not. i'd have to think about it some more.

3. Oct 15, 2005

### Oxymoron

You're right. I don't know what I was thinking.

I have ventured down that avenue. I can show that $f$ maps closed compact sets to closed compact sets.

Im thinking that the easiest way to show that $f(A) = A$ is to show that $f$ is the identity function.

4. Oct 15, 2005

### Hurkyl

Staff Emeritus
Maybe playing with examples would help. You have things like rotations of disks and rings, contractions of intervals, and even the silly map of the unit circle that maps the point with an angle of t to the point with an angle of 2t.

It seems clear that you'll somehow have to actually use the fact X is compact. For example, you should be able to find a map of R that serves as a counterexample.

Last edited: Oct 15, 2005
5. Oct 15, 2005

### Oxymoron

Forget the identity function for now. I have another idea.
Let $A_0 \subseteq X$ be a non-empty closed set such that $f(A_0) \subset A_0$. Then let $A_1 = f(A_0)$ and after iterating we have $f(A_n) \subset A_n$ and

$$A_n = f(A_{n-1}) \quad \forall\, n \in \mathbb{N}$$

From this we see that $\{A_n\}_{n\in\mathbb{N}}^{\infty}$ is a decreasing sequence of non-empty closed sets. Now let

$$A = \bigcap_{n\in\mathbb{N}}A_n$$

So $A$ is also non-empty and closed. If we then observe $f(A)$ we can see that $f(A)$ is certainly contained within $f(A_n)$ and in fact

$$f(A) \subset f(A_n) \backslash A_n$$

for each $n\in\mathbb{N}$. But this simply says that

$$f(A) \subset A$$

Now lets take some point $a \in A$ and let $B = f^{-1}(a)$. Since $a \in A_{n+1} = f(A_n)$ we know that $B \cap A_n \neq \oslash$. Therefore $B \cap A_n$ is a decreasing sequence of non-empty closed sets in a compact space $X$. Therefore

$$\bigcap_{n\in\mathbb{N}} B \cap A_n \neq \oslash$$

Now choose some $y \in \bigcap_{n\in\mathbb{N}} B \cap A_n$. Then obviously $y \in A$ and $f(y) = a$. That is $f(A) = A$.

How does this look?

6. May 9, 2010

### complexnumber

Why is $\{A_n\}_{n\in\mathbb{N}}^{\infty}$ non-empty? Do you need to prove it or is it obvious? I am working on the same problem and cannot figure out this part.