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Let [itex]X[/itex] be a compact Hausdorff space and let [itex]f:X\rightarrow X[/itex] be continuous. Show that there exists a non-empty subset [itex]A \subseteq X[/itex] such that [itex]f(A) = A[/itex].

At the moment I am trying to show that [itex]f[/itex] is a homeomorphism and maybe after that I can show that [itex]f(A) = A[/itex]. But Im not sure if this is the right tactic.

I know that [itex]f[/itex] is a continuous function, and since [itex]f:X \rightarrow X[/itex] it is bijective (?) I thought it might be since the domain and range coincide.

Anyway, if I take a closed subset [itex]A \subseteq X[/itex] then [itex]A[/itex] is automatically compact (since every closed subset of a Hausdorff space is compact). Then since [itex]f[/itex] is continuous it maps closed compact sets to closed compact sets.

Since [itex]f[/itex] is bijective, its inverse [itex]f^{-1}[/itex] exists, and [itex]f^{-1}[/itex] will also map closed compact sets to closed compact sets.

Therefore [itex]f[/itex] is a homeomorphism. Im not sure if I can conclude from this that [itex]A[/itex] is homeomorphic to [itex]f(A)[/itex] which implies [itex]f(A) = A[/itex].

PS. What about the open sets in [itex]X[/itex]? I need help!

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# Homework Help: Compact Hausdorff space with continuous function

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