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As far as I know, there are fixed-point-like results for continuous functions from a convex compact subset $K$ of an Euclidean space to itself. I have one question in mind: Does there exist a set $A\subseteq X$ for which $f(A)=A$? Let's say $X$ is a compact metric space and $f$ is continuous.

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$X = K$? How about $A=\{x\}$ where $x$ is a fixed point? –  Robert Israel Apr 1 '12 at 20:36
Isn't it very trivial Dear sir? X is just a compact metric space. –  El Angel Exterminador Apr 1 '12 at 20:38

2 Answers 2

up vote 4 down vote accepted

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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Thank you dear Sir.I am trying to complete the proof. –  El Angel Exterminador Apr 1 '12 at 20:44
Also, the family of all closed fixed sets (also the family of all fixed sets) is a complete lattice under set inclusion. The smalles element is $\emptyset$ and the largest is the set $A$ constructed by Henno Brandsma in the first paragraph. There also exist minimal nonempty fixed sets by Zorn's lemma, since any chain of nonempty closed fixed sets has a nonempty fixed set that is a lower bound- the intersection of the chain. –  Michael Greinecker Apr 3 '12 at 16:38

Additional Reference: Theorem 1.8 from Dynamical Systems and Ergodic Theory by M. Pollicott and M. Yuri partially answers your question.

Let $T:X\to X$ be a homeomorphism of a compact metric space $X$. Then there exists a non-empty closed set $Y\subset X$ with $TY=Y$ and $T:Y\to Y$ is minimal.

Minimality means that $\{T^ny:n\in\mathbb{Z}\}$ is dense in $Y$ for every $y\in Y$. This result is proved using Zorn's Lemma, much like in Henno Brandsma's answer.

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