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

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The cover of $(0, 1)$ by the intervals $(1 / (k + 2), 1)$ has no Lebesgue number, so the Lebesgue number lemma needs compactness

Statement refuted

Refuted claim: every open cover of a metric space has a Lebesgue number, that is a real $δ > 0$ such that every nonempty subset of diameter less than $δ$ lies inside a single member of the cover.

The true statement carries a compactness hypothesis (Every open cover of a compact metric space has a Lebesgue number: a $δ > 0$ such that every nonempty subset of diameter less than $δ$ lies inside a single member of the cover). The witness is the interval $(0, 1)$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length) as 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), which is 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), covered by

$U_{k} : = (\frac{1}{k + 2}, 1), k \in N .$

For every real $δ > 0$ the interval $A : = (0, t)$ with $t : = \min {δ / 2, 1 / 2}$ is a nonempty subset of $(0, 1)$ of diameter at most $t < δ$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space) that lies inside no $U_{k}$ .

Facts & Assumptions

Given: The interval $(0, 1)$ with the metric $∣ x - y ∣$ restricted to it, and the sets $U_{k} = (1 / (k + 2), 1)$ for $k \in N$ .

[A1]

The refuted claim: every open cover of a metric space has a Lebesgue number.

[L1]

$(0, 1)$ is a metric subspace of $(R, d_{R})$ that is not compact, and the family $(U_{k})_{k \in N}$ consists of sets open in it with union $(0, 1)$ (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 metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, 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]

$diam (A) = \sup {∣ u - v ∣ : u, v \in A}$ for nonempty bounded $A$ , so any upper bound of the distances bounds the diameter (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

[L3]

A Lebesgue number for a cover is a real $δ > 0$ such that every nonempty subset of diameter less than $δ$ lies inside a single member of the cover; a compact metric space has one for every open cover (Every open cover of a compact metric space has a Lebesgue number: a $δ > 0$ such that every nonempty subset of diameter less than $δ$ lies inside a single member of the cover).

[L4]

For every real $η > 0$ there is a natural $m \geq 1$ with $1 / m < η$ , 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).

Counterexample

technique · direct

1.1L1L4

The family $(U_{k})_{k \in N}$ is an open cover of $(0, 1)$ : each $U_{k}$ is the trace on $(0, 1)$ of an open subset of $R$ and is contained in $(0, 1)$ , and any $x \in (0, 1)$ admits a natural $m \geq 1$ with $1 / m < x$ , whence $1 / (m + 2) < 1 / m < x < 1$ and $x \in U_{m}$ .

2.1L2step 1.1

Let $δ > 0$ be real, put $t : = \min {δ / 2, 1 / 2}$ and $A : = (0, t)$ ; then $t > 0$ , $A$ is a nonempty subset of $(0, 1)$ , and every $u, v \in A$ satisfy $∣ u - v ∣ < t$ , so $diam (A) \leq t \leq δ / 2 < δ$ .

3.1L4step 2.1

$A$ is contained in no $U_{k}$ : for a given $k$ the real $x : = \min {t, 1 / (k + 2)} / 2$ satisfies $0 < x \leq t / 2 < t$ , so $x \in A$ , and it satisfies $x \leq (1 / (k + 2)) / 2 < 1 / (k + 2)$ , so $x \notin U_{k}$ .

4.1A1L1L3step 1.1step 2.1step 3.1∎

So no real $δ > 0$ is a Lebesgue number for this cover, and the claim [A1] is refuted; since $(0, 1)$ is not compact, the compactness hypothesis of the Lebesgue number lemma is not removable.

Remarks

What goes wrong. The members of the cover grow towards $(0, 1)$ but none of them reaches down to $0$ , so a set clinging to $0$ of any positive diameter is never captured whole. On a compact space the finitely many members of a subcover put a uniform floor under this, and that floor is the Lebesgue number (Every open cover of a compact metric space has a Lebesgue number: a $δ > 0$ such that every nonempty subset of diameter less than $δ$ lies inside a single member of the cover).

The same cover shows non-compactness directly, having no finite subcover (The open interval $(0, 1)$ is totally bounded and not compact, the cover by the intervals $(1 / (k + 2), 1)$ having no finite subcover), and the failure of uniform continuity of $1 / x$ on $(0, 1)$ is the analytic face of the same phenomenon ( $x \mapsto 1 / x$ is continuous on $(0, 1)$ and not uniformly continuous, so Heine-Cantor needs compactness of the domain).

The cover of $(0, 1)$ by the intervals $(1 / (k + 2), 1)$ has no Lebesgue number, so the Lebesgue number lemma needs compactness

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Depends on

Dependency tree · 2 levels

Sources

The cover of (0,1)(0,1) by the intervals (1/(k+2),1)(1/(k+2),1) has no Lebesgue number, so the Lebesgue number lemma needs compactness

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Depends on

Dependency tree · 2 levels

Sources

The cover of $(0, 1)$ by the intervals $(1 / (k + 2), 1)$ has no Lebesgue number, so the Lebesgue number lemma needs compactness