# Compact Hausdorff space.

1. May 9, 2010

### complexnumber

1. The problem statement, all variables and given/known data

Let $$(X,\tau)$$ be a compact Hausdorff space,
and let $$f : X \to X$$ be continuous, but not surjective. Prove that
there is a nonempty proper subset $$S \subset X$$ such that $$f(S) = S$$. [Hint: Consider the subspaces $$S_n := f^{\circ n}(X)$$ where
$$f^{\circ n} := f \circ \cdots \circ f$$ ($$n$$ times)].

2. Relevant equations

3. The attempt at a solution

If such $$S$$ exists then $$f^{\circ n}(S) = S$$. How should I use this in the proof? I don't have any clue where to start.

2. May 9, 2010

### eok20

What can you say about the sets S_n? For example, are they nested? What do you know about the continuous image of a compact set?

3. Apr 22, 2012

### Flying_Goat

Sorry for digging up an old thread, but I am stuck on the same problem.

I let S = lim S_n so we have f(S) = f(lim S_n) = lim f(S_n) = lim S_{n+1} = S. Obvisouly S is non-empty since each f(S_n) is not empty.

I am not sure if I got it right. We know that each S_n is closed and compact since X is a compact Hausdorff space and f is continuous, but I didnt use this property at all in my solution.

Any help would be appreciated.