We know that:

If $X$ is a metric space, then every contraction has at most one fixed point.

(Note: if metric space is complete, then we have existence and uniqueness)

I wonder if there can be a metric space for which no contraction has a fixed point. Thanks.

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    $\begingroup$ That can't be right: take the rationals and $f (x)=x/2+1/x $. $\endgroup$ – Ian Jul 19 '16 at 3:13
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    $\begingroup$ What you say we know is false: consider the metric space $X=\{2^{-n}:n\in\mathbb N\}$, with the metric induced from $\mathbb R$, wih the map $x\mapsto x/2$. $\endgroup$ – Mariano Suárez-Álvarez Jul 19 '16 at 3:13
  • $\begingroup$ @MarianoSuárez-Alvarez Sorry I mixed up "least" and "most" $\endgroup$ – Shamisen Expert Jul 19 '16 at 3:14
  • $\begingroup$ If it is a complete metric space then by the Banach fixed-point theorem then every contraction must have at least one fixed point. $\endgroup$ – Q the Platypus Jul 19 '16 at 3:21
  • $\begingroup$ You need to define what do you mean by a contraction since there are different definitions. Also, consider only nonconstant contractions and require $X$ to contain more than one point. $\endgroup$ – Moishe Kohan Jul 19 '16 at 3:41

If $X$ is non-empty then we can choose some $x_0\in X$ and define $f(x)=x_0$ for all $x\in X$. This is a contraction with a fixed point.

  • $\begingroup$ The question is looking for a metric space which lacks fixed points on contractions. This answer gives a contraction with a fixed point. $\endgroup$ – Q the Platypus Jul 19 '16 at 3:25
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    $\begingroup$ @QthePlatypus: and therefore shows that no such metric space exists. $\endgroup$ – carmichael561 Jul 19 '16 at 3:32
  • $\begingroup$ You might wish to clarify that you are presenting a counter example. $\endgroup$ – Q the Platypus Jul 19 '16 at 3:45

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