# Can there be a metric space where no contraction has a fixed point?

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.

• That can't be right: take the rationals and $f (x)=x/2+1/x$. – Ian Jul 19 '16 at 3:13
• 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 $x\mapsto x/2$. – Mariano Suárez-Álvarez Jul 19 '16 at 3:13
• @MarianoSuárez-Alvarez Sorry I mixed up "least" and "most" – Shamisen Expert Jul 19 '16 at 3:14
• If it is a complete metric space then by the Banach fixed-point theorem then every contraction must have at least one fixed point. – Q the Platypus Jul 19 '16 at 3:21
• 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. – 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.