Is the following statement true? d(Tx, Ty) < d(x, y) for all x ≠ y on a complete space ⇒ T has a fixed point

Counterexample

Is this true?

d(Tx, Ty) < d(x, y) for all x ≠ y on a complete space ⇒ T has a fixed point

No — it is false.

The counterexample

T(x) = x + 1/x on $[1, \infty)$

Why it fails

Distances strictly decrease but no contraction constant k < 1 exists; no fixed point.

Keep practicing

Finite-dimensional intuition

Something true in ℝⁿ, or in a nice space, assumed in general.

Watch it happen

Baire category, and the argument it powers

Published 23 Aug 2026. Found a mistake? Tell us — corrections are the fastest thing we act on.