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 $x,y\in X$.



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> Suppose $X$ is a metric space such that every contraction map $f:X\to X$ has a unique fixed point. Is $X$ complete?
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Converse to Banach's fixed point theorem?
The answer is no, for example look at the graph of $\sin(1/x)$ on $(0,1]$. But for more information and related questions check out "[On a converse to Banach's Fixed Point Theorem](http://www.ams.org/journals/proc/2009-137-09/S0002-9939-09-09904-3/)" by Márton Elekes.


