# Must a function have a fixed point

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!

• $X=\{-1,1\}$ and $T(x) = -x$? – WimC Feb 4 '17 at 18:42
• 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. – Ann Marie Feb 4 '17 at 18:44
• The finite set of course. – WimC Feb 4 '17 at 18:45
• 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. – Henno Brandsma Feb 4 '17 at 19:01

## 1 Answer

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.