$x\mapsto x+1\mathrm{/}x$ on $[1,\mathrm{\infty })$ strictly decreases every distance and has no fixed point

Refuted claim: a self-map $f$ of a nonempty complete metric space with $d(f(x),f(y))<d(x,y)$ for all $x\mathrm{\ne }y$ has a fixed point; equivalently, the contraction hypothesis of A contraction of a nonempty complete metric space into itself has exactly one fixed point, the limit of the iterates from any starting point may be weakened to that pointwise strict inequality (FALSE: $d(fx,fy)<d(x,y)$ for all $x\mathrm{\ne }y$ on a complete metric space forces a fixed point).

Let $X:=[1,\mathrm{\infty })\subseteq \mathbb{R}$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length) carry the metric $d(x,y):=\mathrm{\mid }x-y\mathrm{\mid }$ inherited from the real line (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ is a metric, its open balls are the intervals $(x-r,x+r)$, and it is unbounded, Isometry, isometric embedding, and the subspace metric on a subset) and let

$f(x):=x+\frac{1}{x}(x\in X)\mathrm{.}$

Then $(X,d)$ is nonempty and complete, $f$ maps $X$ into $X$, $d(f(x),f(y))<d(x,y)$ whenever $x\mathrm{\ne }y$, and $f$ has no fixed point. Moreover $f$ is not a contraction (Lipschitz map, $\alpha$-Hölder map for rational $0<\alpha \le 1$, and contraction): no real $q<1$ satisfies $d(f(x),f(y))\le qd(x,y)$ for all $x,y\in X$.

• The two hypotheses differ by a quantifier, and step 3.1 measures the gap. The shrinking factor at the pair $(n+1,n)$ is $1-1\mathrm{/}(n(n+1))$, which is below $1$ for every $n$ and approaches $1$ as $n$ grows; the contraction condition asks for a single bound below $1$ covering all pairs at once, and that is exactly what fails. The same quantifier move separates continuity from uniform continuity (Uniform continuity of a map of metric spaces: one $\delta$ serving every point).
• Why the iterates do not help. Starting anywhere in $[1,\mathrm{\infty })$, the iterates increase, since $f(x)>x$, and they run off to the right; there is no Cauchy sequence to complete, so completeness of $X$ is no help at all. This is the opposite failure mode to $x\mapsto x\mathrm{/}2$ maps $(0,1]$ into itself, is a $1\mathrm{/}2$-contraction, and has no fixed point, where the iterates are Cauchy and the space is missing their limit.
• Unboundedness is essential to the example, not to the phenomenon as stated here. What this item establishes is only that completeness plus the strict inequality is not enough. It makes no claim about what additional hypothesis would suffice; the classical repair uses compactness, which is a later page of this library.