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Fixed set under continuous map on a compact Hausdorff space

The question we want to answer is as follows. For a nonempty compact Hausdorff topological space X and a continuous function f:X-->X we want to show that there is a fixed set A for f, that is, A is nonempty and f(A)=A.

We also construct an example of a Hausdorff space X which is not compact for which there are no fixed sets, showing that the assumption on compactness is necessary.

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First note that if such A exists then A=f(A) implies f(A)=f(f(A))=f^2(A) and so on: A=f(A)=f^2(A)=f^3(A)=..., and
therefore A equals the intersection of A, f(A), f^2(A).... Here f^2(A)=f(f(A)),f^3(A)=f(f(f(A))) and so on. This is the idea we want to use, here is how:

First note that if B is a compact set in X, then f(B) is also compact since f is continuous. Therefore the sets
X,f(X),f^2(X),...,f^n(X),... are compact in X. Now because X is Hausdorff, those sets
being compact are closed. Note that since f(X) is a subset of X we have that f^{n+1}(X) is a subset f^n(X) so the sets are nested: X contains f(X) which contains f^2(X).... Note also that each set X,f(X),f^2(X),...,f^n(X),... is nonempty. It follows that the
collection of nonempty closed sets {X,f(X),f^2(X),...,f^n(X),...} satisfies the finite intersection property and since
X is compact, their intersection is nonempty

Intersection of all f^n(X), for n=0,1,2,... is nonempty.

The idea would be to define A to be that intersection, A=Intersection of all f^n(X), for n=0,1,2,.... Because the image of the intersection of sets is contained ...

Solution Summary

It is proved that for a compact Hausdorff space X, and a continuous function f on X, there is a fixed set for f, that is, a nonempty subset A of X such that the image of A under f equals A, that is, f(A)=A.

The proof involves the use of Zorn's Lemma. Solution includes a pdf file.

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