show/hide this revision's text 2 I don't think this belongs to general topology -> tags edited
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Converse to Banach's fixed point theorem?

Let $(X,d)$ be a metric space. Banach's fixed point theorem states that if $X$ is complete, then every contraction map $f:X\to X$ has a unique fixed point. A contraction map is a continuous map for which there is an real number $0\leq r < 1$ such that $d(f(x),f(y))\leq rd(x,y)$ holds for all $x,y\in X$.

Suppose $X$ is a metric space such that every contraction map $f:X\to X$ has a unique fixed point. Is $X$ complete?