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If $f:X \rightarrow X$ is continuous and X is compact, will $f$ have a fixed point?

We know that a contraction will have a fixed point but I have not come across an example of a continuous function on a compact set that does not have a fixed point (admittedly I have not worked with functions outside $\mathbb{R}^k$ where Brouwer's fixed point theorem applies).

Is there an example of a continuous function on a compact set such that the function does not have a fixed point?

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4 Answers 4

up vote 6 down vote accepted

Take $X$ to be the unit circle (not disk) and $f$ a non-trivial rotation.

For an example in the real line, take $X=[-2,-1] \cup [1,2]$ and $f(x)=-x$.

What fails in both cases is that $X$ is not convex.

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Let $X=\{-1,1\}$ and let $f(x)=-x$.

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Take the unit circle in $\mathbb{R}^2$ and the map f, as f(x) going to its diametrically opposite point. This map is continuous but has no fixed point.

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Each finite set $X$ is compact. Using the discrete topology on $X$ each map is continuous. So each permutation without a fixpoint will do the job. For $|X|>1$ we always have permutations without a fixpoint.

By the way: The discrete topology is just the induced topology if you consider finite subsets of $\mathbb{R}^n$.

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