Let $X$ be a Hausdorff topological space with the following property:

For every continuous function $f:X\to X$, there is a finite subset $S\neq \emptyset$ of $X$ with $F(S)\subset S$

Does this implies that: either $X$ is a finite set or $X$ has the fixed point property?

What about if $X$ is a manifold? Does the above condition implies FPP for $X$?

Edit: According to the interesting example of Alex to the previous version of question, we ask what about if we require $X$ to be connected?


Answer to the first question: "no"; take $X=\{0\}\cup\{1/n\,|\,n\in\Bbb N\}$.

  • $\begingroup$ very nice example But why you need to two copies?I think one copy is sufficient, not? $\endgroup$ – Ali Taghavi Nov 16 '14 at 20:34
  • $\begingroup$ With one copy it has fixed point property: the only limit point is always fixed. $\endgroup$ – Alex Degtyarev Nov 16 '14 at 20:45
  • 2
    $\begingroup$ I don't understand why, with only one copy, the limit point has to be fixed. Why couldn't $f$ map $1$ to $\frac12$ and map everything else to $1$? $\endgroup$ – Andreas Blass Nov 16 '14 at 21:09
  • $\begingroup$ Ah, sorry, you're right: if the image is finite, the limit point does not need to go to a limit point. Then, indeed, one copy is enough: either $0$ is fixed or image is finite and this image is $S$. $\endgroup$ – Alex Degtyarev Nov 16 '14 at 21:15
  • $\begingroup$ I think it still works (by considering several cases, with images infinite/finite), but one copy is indeed easier, so I've edited. $\endgroup$ – Alex Degtyarev Nov 17 '14 at 7:08

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