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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.

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    this is a special case of what is known as the Banach fixed point theorem or the contraction mapping theorem (note that for metric spaces compact$\implies$ complete) – 2017-01-04

1 Answers 1

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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 $f$.