Contraction mapping in an incomplete metric space Let us consider a contraction mapping $f$ acting on metric space $(X,~\rho)$ ($f:X\to X$ and for any $x,y\in X:\rho(f(x),f(y))\leq k~\rho(x,y),~ 0 < k < 1$). If $X$ is complete, then there exists an unique fixed point. But is there an incomplete space for which this property holds as well? I think $X$ should be something like graph of $\text{sin}~{1\over x}$, but I don't know how to prove it.
Thank you and sorry for my english.
 A: There is a sense in which every such fixed point comes from a complete metric space:
Theorem: Let $X$ be an arbitrary set. Let $f:X\to X$ be a function with fixed point $x$ such that for all positive natural numbers $n$ and for all $y\neq x$, we have $f^n(y)\neq y$. Then there exists a complete metric $d$ on $X$ such that $f$ is a strict contraction.
The theorem is originally due to C. Bessaga and is not that easy to prove. A relatively short proof can be found here.
A: Section 4 of this paper of Suzuki and Takahashi gives an example of a metric space -- in fact, a subspace of the Euclidean plane (so it seems you were on the right track!) -- which is incomplete but for which every contraction mapping has a fixed point.
They go on to repair matters by defining a "weakly contractive mapping" and showing that a metric space is complete iff every weakly contractive mapping has a fixed point.
Note: I was not aware of this paper until I read this question.  I then googled --
contraction mapping, characterization of completeness -- and the paper showed up right away.  (I look forward to reading it more carefully when I get the chance...)
