= Strict distance decrease on a compact metric space
{title2=$d(fx,fy)<d(x,y)\quad(x\ne y)$}
A self-map $f$ of a nonempty <compact metric space> satisfying the displayed strict distance inequality has exactly one <fixed point>, although a uniform <contraction> constant need not exist. The inequality makes $f$ <Lipschitz continuous> with constant at most one, so $g(x)=d(fx,x)$ is continuous and attains a minimum at $p$. If $fp\ne p$, then $g(fp)<g(p)$, a contradiction. Two different <fixed points> contradict the same distance inequality.
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