A question about topology

Question

Assume $(X,d)$ is a compact metric space,

$f:X\rightarrow X$, if $x\neq y$，then $d(f(x),f(y))＜d(x,y)$

prove that there exists $x$，such that $f(x)＝x$

Thanks for your help.

Answer 1

HINT: Let $K_0=X$, and for $n\in\Bbb N$ let $K_{n+1}=f[K_n]$.

• Show that $K_{n+1}\subseteq K_n$ for each $n\in\Bbb N$.
• Show that $\operatorname{diam}K_{n+1}<\operatorname{diam}K_n$ for each $n\in\Bbb N$, where $\operatorname{diam}K=\sup\{d(x,y):x,y\in K\}$.

Let $K=\bigcap_{n\in\Bbb N}K_n$

• Show that $K\ne\varnothing$ and that $f[K]=K$.
• Conclude that $K=\{x\}$ for some fixed point $x$ of $f$.