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If $\omega:H \rightarrow H$ (into) is continuous and $H$ is compact, does $\omega$ fix any of its subsets other than the empty set?

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2 Answers 2

up vote 2 down vote accepted

the proof I learned from Henno Brandsma's answer: Let $X$ be compact Hausdorff (no metric is needed), and define $A_0 = X$, $A_{n+1} = f[A_n]$; then all $A_n$ are compact non-empty, and the $A_n$ are decreasing. Try to show that $A = \cap_n A_n$, which is also compact and non-empty, satisfies $f[A] = A$.

Another non-constructive way to show this is to consider the poset $\mathcal{P} = \{ A \subset X \mid A, \mbox{closed, non-empty and } f[A] \subset A \}$, ordered under reverse inclusion. Then an upper bound for a chain from $\mathcal{P}$ is the (non-empty) intersection, and a maximal element (by Zorn one exists) is a set $A$ with $f[A] = A$.

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That's a good proof, but I don't need X (or H) Hausdorff. –  cap Apr 25 '12 at 6:05
    
I assume you are in $\mathbb{R}$ and you already said your H is compact, do you know $\mathbb{R}$ is hausdorff space? –  Une Femme Douce Apr 25 '12 at 6:13
    
H is simply a metric space; no assumptions of R. –  cap Apr 25 '12 at 6:15
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do you know how to prove any metric space is hausdorff? –  Une Femme Douce Apr 25 '12 at 6:16
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$A$ will be all of $X$ whenever $f$ is surjective. –  Chris Eagle Apr 25 '12 at 6:28

In general, the answer is no. For example, let $H$ be a two-point space and $\omega$ be the function that swaps the two points.

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