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Let $(X, \rho)$ be a compact metric space.

Suppose that $T: X \rightarrow X$ and for all $u \neq v \in X$,

$$ \rho(T(u),T(v)) \leq \rho(u,v) $$

Must $T$ have a fixed point?

My intuition is telling me that the answer is no, since the contraction mapping principle requires

$$ \rho(T(u),T(v)) \leq c\rho(u,v) $$

with $c \in (0,1)$ for there to be a unique fixed point. However, I am struggling to come up with a counterexample.

Thank you in advance for the help!

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    $\begingroup$ $X=\{-1,1\} $ and $T(x) = -x$? $\endgroup$ – WimC Feb 4 '17 at 18:42
  • $\begingroup$ Do you mean the finite set of just $-1$ and $1$ or the interval $[-1,1]$? If it is the interval, then it has a fixed point at 0. $\endgroup$ – Ann Marie Feb 4 '17 at 18:44
  • $\begingroup$ The finite set of course. $\endgroup$ – WimC Feb 4 '17 at 18:45
  • $\begingroup$ It's not intuition, but sort of a meta-feeling: if they could have proved it for $c=1$, you would have learnt it as a theorem, so they're probably looking for a counterexample. $\endgroup$ – Henno Brandsma Feb 4 '17 at 19:01
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You may take the circle, $S^1={\Bbb R}/{\Bbb Z}$ and an irrational rotation $T:\phi\mapsto \phi + \alpha$ with $\alpha\in {\Bbb R}\setminus {\Bbb Q}$. It has no fixed point and no periodic orbits.

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