Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

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 $f[A] = A$.

share|improve this answer
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
do you know how to prove any metric space is hausdorff? –  Une Femme Douce Apr 25 '12 at 6:16
$A$ 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.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.