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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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 $\delta > 0$ such that every nonempty subset of diameter less than $\delta$ 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 $\delta > 0$ such that every nonempty subset of diameter less than $\delta$ lies inside a single member of the cover). The witness is the interval $(0,1)$ (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length) as a metric subspace of $\mathbb{R}$ (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), 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

$U_k \;:=\; \Big(\tfrac{1}{k+2},\ 1\Big), \qquad k \in \mathbb{N}.$

For every real $\delta > 0$ the interval $A := (0,\ t)$ with $t := \min\{\delta/2,\ 1/2\}$ is a nonempty subset of $(0,1)$ of diameter at most $t < \delta$ (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 $k \in \mathbb{N}$ .

[A1]

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

[L1]

$(0,1)$ is a metric subspace of $(\mathbb{R},d_{\mathbb{R}})$ that is not compact, and the family $(U_k)_{k \in \mathbb{N}}$ consists of sets open in it with union $(0,1)$ (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, Intervals of $\mathbb{R}$ : 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]

$\operatorname{diam}(A) = \sup\{|u-v| : u,v \in A\}$ 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 $\delta > 0$ such that every nonempty subset of diameter less than $\delta$ 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 $\delta > 0$ such that every nonempty subset of diameter less than $\delta$ lies inside a single member of the cover).

[L4]

For every real $\eta > 0$ there is a natural $m \ge 1$ with $1/m < \eta$ , and reciprocals of positives are positive and reverse the order (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, Inverses of positives are positive, and reciprocation reverses order).

Counterexample

technique · direct

1.1

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

L1L4

2.1

Let $\delta > 0$ be real, put $t := \min\{\delta/2,\ 1/2\}$ 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 $\operatorname{diam}(A) \le t \le \delta/2 < \delta$ .

L2step 1.1

3.1

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

L4step 2.1

4.1

So no real $\delta > 0$ 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.

A1L1L3step 1.1step 2.1step 3.1∎

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 $\delta > 0$ such that every nonempty subset of diameter less than $\delta$ 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).

Depends on

Used by

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Sources

Lebesgue's number lemma (Wikipedia) (standard reference, not scraped)
Compact space (Wikipedia) (standard reference, not scraped)

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