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 not uniformly continuous, so Heine-Cantor needs compactness of the domain

Statement refuted

Refuted claim: a continuous map from a bounded metric space to a metric space is uniformly continuous.

The true statement replaces bounded by compact (Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous). The witness is $g (x) = 1 / x$ on the interval $(0, 1)$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length), a metric subspace of $R$ (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) that is bounded and not compact (The open interval $(0, 1)$ is totally bounded and not compact, the cover by the intervals $(1 / (k + 2), 1)$ having no finite subcover). The map is continuous (Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form) and is not uniformly continuous (Uniform continuity of a map of metric spaces: one $δ$ serving every point): the pairs

$x_{k} : = \frac{1}{k + 2}, y_{k} : = \frac{1}{k + 3} (k \in N)$

lie in $(0, 1)$ , satisfy $∣ x_{k} - y_{k} ∣ \to 0$ , and yet $∣ g (x_{k}) - g (y_{k}) ∣ = 1$ for every $k$ . The indices are written $k + 2$ and $k + 3$ because $N$ contains $0$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences), so that both points lie in $(0, 1)$ already at $k = 0$ .

Facts & Assumptions

Given: The interval $(0, 1)$ with the metric $∣ x - y ∣$ restricted to it, the map $g (x) = 1 / x$ on it, and the points $x_{k} = 1 / ι (k + 2)$ and $y_{k} = 1 / ι (k + 3)$ .

[A1]

The refuted claim: a continuous map from a bounded metric space to a metric space is uniformly continuous.

[L1]

$(0, 1)$ is a bounded metric subspace of $(R, d_{R})$ and is not compact (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, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, The open interval $(0, 1)$ is totally bounded and not compact, the cover by the intervals $(1 / (k + 2), 1)$ having no finite subcover, Open cover, subcover, compact metric space, and compact subset of a metric space, Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[L2]

$g$ is continuous at $c$ when for every real $ε > 0$ there is a real $δ > 0$ with $∣ g (x) - g (c) ∣ < ε$ whenever $∣ x - c ∣ < δ$ ; it is uniformly continuous when one $δ$ works for all pairs at once (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).

[L3]

For every real $η > 0$ there is a natural $m \geq 1$ with $1 / m < η$ ; canonical naturals are positive and increasing, and reciprocals of positives are positive and reverse the order (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, The canonical natural $ι (n) = n \cdot 1_{F}$ of a field).

[L4]

A continuous map from a compact metric space to a metric space is uniformly continuous (Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous).

Counterexample

technique · direct

1.1L2L3

$g$ is continuous at each $c \in (0, 1)$ : given a real $ε > 0$ , the choice $δ : = \min {c / 2, ε c^{2} / 2}$ gives, for $∣ x - c ∣ < δ$ with $x \in (0, 1)$ , first $x > c / 2 > 0$ and then $∣ 1 / x - 1 / c ∣ = ∣ c - x ∣ / (x c) < (2 / c^{2}) ∣ x - c ∣ < ε$ .

2.1L3step 1.1

For every $k \in N$ the points $x_{k} = 1 / ι (k + 2)$ and $y_{k} = 1 / ι (k + 3)$ lie in $(0, 1)$ , since $ι (k + 2) \geq 2 > 1$ and $ι (k + 3) \geq 3 > 1$ make both reciprocals positive and below $1$ .

3.1L3step 2.1

$∣ g (x_{k}) - g (y_{k}) ∣ = ∣ ι (k + 2) - ι (k + 3) ∣ = 1$ for every $k$ , because $ι (k + 3) = ι (k + 2) + 1$ .

3.2L3step 2.1

$∣ x_{k} - y_{k} ∣ = 1 / ι (k + 2) - 1 / ι (k + 3) < 1 / ι (k + 2)$ , and given a real $δ > 0$ a natural $m \geq 1$ with $1 / m < δ$ gives $∣ x_{m} - y_{m} ∣ < 1 / ι (m + 2) < 1 / m < δ$ .

4.1L2step 3.1step 3.2

So no real $δ > 0$ witnesses uniform continuity at $ε = 1$ : by step 3.2 some pair $x_{m}, y_{m}$ of points of $(0, 1)$ has $∣ x_{m} - y_{m} ∣ < δ$ , while step 3.1 gives $∣ g (x_{m}) - g (y_{m}) ∣ = 1$ , which is not less than $1$ .

5.1A1L1L4step 1.1step 4.1∎

Hence $g$ is a continuous map on the bounded, non-compact space $(0, 1)$ that is not uniformly continuous, and the claim [A1] is refuted: the compactness hypothesis of Heine-Cantor cannot be weakened to boundedness.

Remarks

Where the argument would break on a compact domain. On $[0, 1]$ the same pairs converge to $0$ , and any continuous function there is uniformly continuous [L4]; the map $1 / x$ escapes that only because $0$ is missing from its domain, so the values are free to run away as the arguments approach the missing point.

The failure is exactly the failure of a Lebesgue number. The proof of Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous turns pointwise continuity into uniform continuity by producing one radius that works everywhere, and the cover of $(0, 1)$ used in The cover of $(0, 1)$ by the intervals $(1 / (k + 2), 1)$ has no Lebesgue number, so the Lebesgue number lemma needs compactness shows that no such radius exists here.

$x \mapsto 1 / x$ is continuous on $(0, 1)$ and not uniformly continuous, so Heine-Cantor needs compactness of the domain

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 not uniformly continuous, so Heine-Cantor needs compactness of the domain

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Depends on

Dependency tree · 2 levels

Sources

$x \mapsto 1 / x$ is continuous on $(0, 1)$ and not uniformly continuous, so Heine-Cantor needs compactness of the domain