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

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$x \mapsto 1 / x$ is continuous on $(0, 1)$ and sends the Cauchy sequence $(1 / (k + 2))_{k \geq 0}$ to an unbounded one

Statement refuted

Refuted claim: the hypothesis of A uniformly continuous map sends Cauchy sequences to Cauchy sequences may be weakened from uniform continuity to continuity; a continuous map of metric spaces sends Cauchy sequences to Cauchy sequences.

Let $X : = (0, 1) \subseteq R$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length) with the metric 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), let $R$ carry its usual metric, and let $f : X \to R$ be $f (x) : = 1 / x$ . Then $f$ is continuous (Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form), the sequence $x_{k} : = 1 / (k + 2)$ is Cauchy in $X$ (Cauchy sequence in a metric space), and the image sequence $f (x_{k}) = k + 2$ is unbounded and not Cauchy. Consequently $f$ is not uniformly continuous (Uniform continuity of a map of metric spaces: one $δ$ serving every point).

Facts & Assumptions

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

[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 $x, a > 0$ : $∣ 1 / x - 1 / a ∣ = ∣ x - a ∣ / (x a)$ ; reciprocation reverses order on the positives; and inequalities may be multiplied by positives (Inverses of positives are positive, and reciprocation reverses order, Sign rules for products and monotonicity of multiplication).

[L3]

For every real $η > 0$ there is a natural $N \geq 1$ with $1 / N < η$ , and for every real $t$ there is a natural $n \geq 1$ with $t < n$ (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).

[L4]

Two reals have a minimum (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

[L5]

Continuity at a point, uniform continuity, Cauchyness and convergence, all testable with real $ε > 0$ (Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form, Uniform continuity of a map of metric spaces: one $δ$ serving every point, 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).

[L6]

A convergent sequence in a metric space is Cauchy (Every convergent sequence in a metric space is Cauchy); a sequence of reals is bounded when some real dominates all its absolute values (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L7]

Uniformly continuous maps send Cauchy sequences to Cauchy sequences (A uniformly continuous map sends Cauchy sequences to Cauchy sequences).

Counterexample

technique · direct

1.1L1L2L4L5

$f$ is continuous at every $a \in X$ : put $δ : = \min {a / 2, ε a^{2} / 4} > 0$ . If $x \in X$ and $∣ x - a ∣ < δ$ then $∣ x - a ∣ < a / 2$ , so $x > a / 2$ , and therefore $∣ 1 / x - 1 / a ∣ = ∣ x - a ∣ / (x a) < δ / ((a / 2) a) = 2 δ / a^{2} \leq ε / 2 < ε$ . Since $a$ was arbitrary, $f$ is continuous on $X$ .

1.2L1L2

Every term of $(x_{k})$ lies in $X$ : $k + 2 \geq 2$ gives $0 < 1 / (k + 2) \leq 1 / 2 < 1$ .

2.1step 1.2L1L3L5

$(x_{k})$ is Cauchy in $X$ : given a real $ε > 0$ , take $N \geq 1$ with $1 / N < ε / 2$ ; for $k, l \geq N$ we have $k + 2 > N$ and $l + 2 > N$ , so $∣ x_{k} - x_{l} ∣ \leq 1 / (k + 2) + 1 / (l + 2) < 2 / N < ε$ .

2.2step 1.2L2

$f (x_{k}) = k + 2$ for every $k$ .

3.1step 2.2L3L6

The image sequence is unbounded: for a real $M$ , [L3] supplies a natural $n \geq 1$ with $M < n$ , and then $f (x_{n}) = n + 2 > n > M$ .

3.2step 2.2L5

The image sequence is not Cauchy: $∣ f (x_{K}) - f (x_{K + 1}) ∣ = ∣ (K + 2) - (K + 3) ∣ = 1$ for every $K$ , so the Cauchy condition fails at $ε = 1$ .

4.1step 1.1step 2.1step 3.1step 3.2L7∎

So a continuous map has carried a Cauchy sequence to a non-Cauchy one, which refutes the claim above; and $f$ cannot be uniformly continuous, since a uniformly continuous map would have preserved Cauchyness.

Remarks

Where the $δ$ escapes. The $δ$ produced in step 1.1 is proportional to $a^{2}$ , so it shrinks to nothing as $a$ approaches $0$ ; there is no single $δ$ serving every point, which is exactly the failure of uniform continuity (Uniform continuity of a map of metric spaces: one $δ$ serving every point). The Cauchy sequence $(x_{k})$ walks into the region where the $δ$ s vanish.
The domain, not the formula, is the problem. On $[a, 1)$ with $a > 0$ the same map is Lipschitz with constant $1 / a^{2}$ , hence uniformly continuous, and it preserves Cauchy sequences there. It is the missing endpoint $0$ of $(0, 1)$ that makes the example work, and that is the same missing point as in FALSE: every Cauchy sequence in a metric space converges.
Indexing. The sequence is $1 / (k + 2)$ and not $1 / k$ , because $N$ contains $0$ here (Sequences of reals: bounded, eventually, frequently, tails, subsequences) and $1 / 0$ does not exist; and not $1 / (k + 1)$ , because that equals $1$ at $k = 0$ and $1 \notin (0, 1)$ .
This is one of the two witnesses named in Contraction implies Lipschitz implies uniformly continuous implies continuous; every Hölder map is uniformly continuous, and a Lipschitz map on a bounded space is Hölder for every exponent, the one separating continuity from uniform continuity. The other, separating Hölder from Lipschitz, is $\sqrt{\cdot}$ on $[0, \infty)$ is uniformly continuous and exactly $1 / 2$ -Hölder, and is not Lipschitz.

$x \mapsto 1 / x$ is continuous on $(0, 1)$ and sends the Cauchy sequence $(1 / (k + 2))_{k \geq 0}$ to an unbounded one

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Depends on

Dependency tree · 2 levels

Sources

x↦1/xx↦1/x is continuous on (0,1)(0,1) and sends the Cauchy sequence (1/(k+2))k≥0(1/(k+2))k≥0​ to an unbounded one

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Depends on

Dependency tree · 2 levels

Sources

$x \mapsto 1 / x$ is continuous on $(0, 1)$ and sends the Cauchy sequence $(1 / (k + 2))_{k \geq 0}$ to an unbounded one