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$X$ is a compact metric space, $f$ is a continuous function from $X$ $\rightarrow$ $X$, which of the following must be true?

  A. $f$ has fixed point

  B. $f$ is a closed map

  C. $f$ is uniformly continuous

I know C is right by compactness. What about the others?

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2 Answers 2

up vote 12 down vote accepted

Hint:

  • For A, consider a two-element set $\{x,y\}$ with the discrete metric.

  • For B, use that closed subsets of compact spaces are compact, that compact subsets of Hausdorff spaces are closed, and that the continuous image of a compact set is compact.

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So B is right?As any closed set in compact metric space is also compact. –  Jebei Apr 8 '13 at 11:02
    
Yes, B is right, though you would probably want to expand a bit on the argument to get full credit for the solution. –  Zev Chonoles Apr 8 '13 at 11:04

For a non-discrete counterexample to A, consider $X=S^1$ and think about rotations. More generally, any nontrivial rotation of $\mathbb R^n$, with $n$ odd, about the origin has only the origin as a fixed point. So take $X=\mathbb D_1 -D_2$ with D_1 a closed disk about the origin and $D_2$ an open disk of smaller radius about the origin.

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but $\mathbb R^n$-$\{0\}$ isn't compact anymore, even if you replace $\mathbb R^n$ by $D^n$. –  Stefan Hamcke Apr 8 '13 at 11:11
3  
@Ittay: I think that in order for rotations of $S^n$ not to have fixed points, you need $n$ to be odd (or, technically, $n=0$ also works, which is really the same thing as my example of a two-point discrete set). –  Zev Chonoles Apr 8 '13 at 11:15
    
Thank you both for the remarks, I made some correction. –  Ittay Weiss Apr 8 '13 at 11:17
    
@ZevChonoles I like your observation that this is really your example in higher dimensions. –  Ittay Weiss Apr 8 '13 at 11:18
    
@IttayWeiss Why n must be odd? –  Jebei Apr 8 '13 at 13:57

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