CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck 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 $(1/(k+2))_{k \ge 0}$ 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 $X := (0,1) \subseteq \mathbb{R}$ (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length) with the metric inherited from the real line (The absolute value makes $\mathbb{R}$ 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 $\mathbb{R}$ carry its usual metric, and let $f : X \to \mathbb{R}$ be $f(x) := 1/x$ . Then $f$ is continuous (Continuity of a map between metric spaces, at a point and globally, in the $\varepsilon$ - $\delta$ 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 $\delta$ serving every point).

Facts & Assumptions

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

[L1]

The absolute value makes $\mathbb{R}$ a metric space, and a restriction of a metric to a subset is a metric with the same distances (The absolute value makes $\mathbb{R}$ 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 $x,a > 0$ : $|1/x - 1/a| = |x-a|/(xa)$ ; 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 $\eta > 0$ there is a natural $N \ge 1$ with $1/N < \eta$ , and for every real $t$ there is a natural $n \ge 1$ with $t < n$ (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ , 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 $\varepsilon > 0$ (Continuity of a map between metric spaces, at a point and globally, in the $\varepsilon$ - $\delta$ form, Uniform continuity of a map of metric spaces: one $\delta$ 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 $\mathbb{R}$ , 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

1.1

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

L1L2L4L5

1.2

Every term of $(x_k)$ lies in $X$ : $k+2 \ge 2$ gives $0 < 1/(k+2) \le 1/2 < 1$ .

L1L2

2.1

$(x_k)$ is Cauchy in $X$ : given a real $\varepsilon > 0$ , take $N \ge 1$ with $1/N < \varepsilon/2$ ; for $k,l \ge N$ we have $k+2 > N$ and $l+2 > N$ , so $|x_k - x_l| \le 1/(k+2) + 1/(l+2) < 2/N < \varepsilon$ .

step 1.2L1L3L5

2.2

$f(x_k) = k+2$ for every $k$ .

step 1.2L2

3.1

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

step 2.2L3L6

3.2

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 $\varepsilon = 1$ .

step 2.2L5

4.1

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.

step 1.1step 2.1step 3.1step 3.2L7∎

Remarks

Where the $\delta$ escapes. The $\delta$ produced in step 1.1 is proportional to $a^2$ , so it shrinks to nothing as $a$ approaches $0$ ; there is no single $\delta$ serving every point, which is exactly the failure of uniform continuity (Uniform continuity of a map of metric spaces: one $\delta$ serving every point). The Cauchy sequence $(x_k)$ walks into the region where the $\delta$ 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 $\mathbb{N}$ 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.

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Sources

Uniform continuity (Wikipedia) (standard reference, not scraped)
Cauchy sequence (Wikipedia) (standard reference, not scraped)

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