Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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.

share|cite|improve this answer

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

share|cite|improve this answer

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.

share|cite|improve this answer

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 $\mathbb{R}^n$.

share|cite|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.