CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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$x \mapsto x / 2$ maps $(0, 1]$ into itself, is a $1 / 2$ -contraction, and has no fixed point

Statement refuted

Refuted claim: the completeness hypothesis in Banach's fixed point theorem (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 dropped; a contraction of a nonempty metric space into itself always has a fixed point.

Let $X : = (0, 1] \subseteq R$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length) carry the metric $d (x, y) : = ∣ x - y ∣$ inherited from the real line (The absolute value makes $R$ a metric space: $d (x, y) = ∣ x - y ∣$ 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 $f : X \to X$ be $f (x) : = x / 2$ . Then $X$ is nonempty, $f$ maps $X$ into itself and is a contraction with constant $1 / 2$ (Lipschitz map, $α$ -Hölder map for rational $0 < α \leq 1$ , and contraction), and $f$ has no fixed point in $X$ . The single 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 that fails is completeness (Complete metric space: every Cauchy sequence converges in the space), and it does fail.

Facts & Assumptions

Given: The interval $X = (0, 1]$ with the metric $d$ inherited from $R$ ; the map $f (x) = x / 2$ ; the sequence $x_{k} : = 1 / (k + 2)$ ; a real $ε > 0$ .

[L1]

The absolute value makes $R$ a metric space, and a restriction of a metric to a subset is a metric with the same distances (The absolute value makes $R$ a metric space: $d (x, y) = ∣ x - y ∣$ 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, Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric, Basic properties of the absolute value).

[L2]

For every real $η > 0$ there is a natural $N \geq 1$ with $1 / N < η$ ; and $0 < a < b$ gives $0 < 1 / b < 1 / a$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , Every complete ordered field is Archimedean, Inverses of positives are positive, and reciprocation reverses order).

[L3]

A convergent sequence in a metric space is Cauchy, limits are unique, and both notions may be tested with real $ε > 0$ (Every convergent sequence in a metric space is Cauchy, A sequence in a metric space has at most one limit, Cauchy sequence in a metric space, Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ , The rationals embed densely in the reals).

[L4]

Contraction: Lipschitz with a constant $q$ satisfying $0 \leq q < 1$ (Lipschitz map, $α$ -Hölder map for rational $0 < α \leq 1$ , and contraction).

Counterexample

technique · direct

1.1L1

$X$ is nonempty, since $1 \in X$ .

1.2L1L2

$f$ maps $X$ into $X$ : for $0 < x \leq 1$ one has $0 < x / 2 \leq 1 / 2 \leq 1$ .

1.3L1L4

$f$ is a contraction with constant $1 / 2$ : $d (f (x), f (y)) = ∣ x / 2 - y / 2 ∣ = \frac{1}{2} ∣ x - y ∣ = \frac{1}{2} d (x, y)$ for all $x, y \in X$ , and $0 \leq 1 / 2 < 1$ .

1.4L1L2L3

Every term of $(x_{k})$ lies in $X$ , since $k + 2 \geq 2$ gives $0 < 1 / (k + 2) \leq 1 / 2 \leq 1$ ; and $x_{k} \to 0$ in $R$ , because for a real $ε > 0$ and $N \geq 1$ with $1 / N < ε$ every $k \geq N$ has $k + 2 > N$ and hence $∣ x_{k} - 0 ∣ = 1 / (k + 2) < 1 / N < ε$ .

2.1step 1.2L1

$f$ has no fixed point in $X$ : $f (x) = x$ means $x / 2 = x$ , hence $x = 0$ , and $0 \notin X$ .

2.2step 1.4L1L3

So $(x_{k})$ is Cauchy in $R$ , hence Cauchy in $(X, d)$ , the distances being the same; and it has no limit in $X$ , since a limit $p \in X$ would also be a limit in $R$ and uniqueness of limits there would force $p = 0 \notin X$ . Hence $(X, d)$ is not complete.

3.1step 1.1step 1.3step 2.1step 2.2∎

Therefore $X$ is a nonempty metric space and $f$ a contraction of it into itself with no fixed point, so the completeness 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 cannot be dropped.

Remarks

The missing point is exactly the fixed point. In $R$ the map $x \mapsto x / 2$ has the fixed point $0$ , and the iterates $x_{n} = x_{0} / 2^{n}$ from any starting point in $(0, 1]$ converge to $0$ . They are Cauchy in $(0, 1]$ , as Banach's proof guarantees, and the space simply has nowhere to put their limit. So the theorem's proof runs correctly up to the last step, and completeness is precisely what that last step needs.
Closedness would fix it. $[0, 1]$ is closed in $R$ , hence complete (Closed subspaces of complete metric spaces are complete; the converse under countable choice, $R$ and $R^{n}$ for $n \geq 1$ with the Euclidean metric are complete, componentwise from the Cauchy criterion in $R$ ), and the same map on it has the fixed point $0$ . This is the usual way the hypothesis is met in practice.
Every other hypothesis is present, which is what makes this a clean counterexample rather than a curiosity: $X$ is nonempty, $f$ maps $X$ into $X$ , and the contraction constant $1 / 2$ is explicit and uniform.

$x \mapsto x / 2$ maps $(0, 1]$ into itself, is a $1 / 2$ -contraction, and has no fixed point

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Depends on

Dependency tree · 2 levels

Sources

x↦x/2x↦x/2 maps (0,1](0,1] into itself, is a 1/21/2-contraction, and has no fixed point

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Depends on

Dependency tree · 2 levels

Sources

$x \mapsto x / 2$ maps $(0, 1]$ into itself, is a $1 / 2$ -contraction, and has no fixed point