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Can someone provide an example to show that for the Banach fixed point theorem, that is if $T : X → X$ is a contraction in a complete metric space $(X, d)$ then $T$ has a unique fixed point that $X$ is complete is an essential condition?

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  • $\begingroup$ It is not necessary for $X$ to be complete. Take $T: \mathbb{Q} \to \mathbb{Q}$ defined by $T(q) = q/2$, where $\mathbb{Q}$ is the rational numbers with the usual metric. $\mathbb{Q}$ is not complete. $T$ is a contraction with a unique fixed point ($0$). Did I misunderstand something? Maybe you are asking why the fixed point might not exist if $X$ is not complete? The answer below is an example. $\endgroup$ – Stefan Smith Oct 18 '13 at 2:49
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$$T \colon \mathbb{R}\setminus\{0\} \to \mathbb{R}\setminus\{0\};\; T(x) = x/2.$$

Basically, all examples are of that form, since a contraction is uniformly continuous, and can thus be extended to the completion of $X$, and the extension is also a contraction. So the extension $\widehat{T}\colon \widehat{X}\to\widehat{X}$ satisfies the premises of Banach's fixed point theorem, and has a unique fixed point.

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