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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.

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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...)

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    $\begingroup$ That's a real interesting paper. The counterexample in the paper is the following: For each number $n=1,2,3,...$, let $L_n$ be the line segment going from $(0,0)$ to $(1,1/n)$ (including the end points). Union all the $L_n$ together to get the desired space. Incidentally, the proof that this space is a counterexample is also quite easy. $\endgroup$ – Jason DeVito Sep 28 '11 at 19:13
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    $\begingroup$ Incidentally, that space is similar to Number 119 in Steen and Seebach's Counterexamples in Topology, which they call "the infinite broom". $\endgroup$ – Nate Eldredge Sep 28 '11 at 21:50
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    $\begingroup$ The link to that article is now broken. Working link: Suzuki, T., Takahashi, W.: Fixed point Theorems and characterizations of metric completeness. Thanks! $\endgroup$ – Weaam Oct 27 '15 at 20:53
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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.

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    $\begingroup$ +1. I first learned of this converse to CMP in a talk Keith Conrad gave for our Undergraduate Math Club. As usual, he has a very nice writeup, math.uconn.edu/~kconrad/blurbs/analysis/contractionshort.pdf, although (somewhat unusually) he does not give the proof of Bessaga's Theorem: as you say, it's not so easy. I also found, and liked, the paper by Jachymski that you link to (it contains a variant of Bessaga's Theorem that I independently noticed and suggested to Keith after hearing his talk). $\endgroup$ – Pete L. Clark Jan 18 '12 at 13:32
  • $\begingroup$ I have it from the Handbook of Analysis and Its Foundations by Schechter, which I cherish. It also contains a different converse, due to Meyers. The statement can be glanced at here. $\endgroup$ – Michael Greinecker Jan 18 '12 at 17:06
  • $\begingroup$ (I didn't get pinged...) I agree, that's a really good book. $\endgroup$ – Pete L. Clark Apr 18 '15 at 21:08

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