Non-Surjective Continuous Function in Compact Hausdorff Space?

[complexnumber]

Homework Statement

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) =<br /> S. [Hint: Consider the subspaces S_n := f^{\circ n}(X) where
f^{\circ n} := f \circ \cdots \circ f (n times)].

Homework Equations

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.

 

[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?

 

[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.