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

1.1

$g$ 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 $∣ 1 / x - 1 / c ∣ = ∣ c - x ∣ / (x c) < (2 / c^{2}) ∣ x - c ∣ < ε$ .

L2L3

2.1

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 $1$ .

L3step 1.1

3.1

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

L3step 2.1

3.2

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

L3step 2.1

4.1

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 $1$ .

L2step 3.1step 3.2

5.1

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.

A1L1L4step 1.1step 4.1∎

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.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

50 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Heine-Cantor theorem (Wikipedia) (standard reference, not scraped)
Uniform continuity (Wikipedia) (standard reference, not scraped)

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