ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced — the exact statement appears in a cited source; only wording and notation differ.
AI-adapted — a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated — a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The cover of $[0,1]$ by $(-1, 2/3)$ and $(1/3, 2)$ has Lebesgue number $1/3$ , and no larger one

Example

Work in $\mathbb{R}$ with its usual metric $d(x,y) = |x-y|$ (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) and let $[0,1]$ carry the restricted metric (Isometry, isometric embedding, and the subspace metric on a subset, Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length). Put

$U := (-1,\ 2/3) \cap [0,1] = [0,\ 2/3), \qquad V := (1/3,\ 2) \cap [0,1] = (1/3,\ 1],$

so that $\{U,V\}$ is an open cover of the compact metric space $[0,1]$ (Open cover, subcover, compact metric space, and compact subset of a metric space, Heine-Borel in $\mathbb{R}^n$ : with the Euclidean metric a subset of $\mathbb{R}^n$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line). Then:

$\delta = 1/3$ is a Lebesgue number for $\{U,V\}$ (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): every nonempty $A \subseteq [0,1]$ with $\operatorname{diam}(A) < 1/3$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space) is contained in $U$ or in $V$ .
No larger $\delta$ works: for every real $\delta' > 1/3$ the set $A := [1/3,\ 2/3]$ is nonempty with $\operatorname{diam}(A) = 1/3 < \delta'$ and is contained in neither $U$ nor $V$ .

Facts & Assumptions

Given: The compact metric space $[0,1]$ with $d(x,y) = |x-y|$ , and the sets $U = [0,2/3)$ and $V = (1/3,1]$ .

[L1]

A closed bounded subset of $\mathbb{R}$ is a compact subset of $(\mathbb{R},d)$ , and $[0,1]$ is closed and bounded (Heine-Borel in $\mathbb{R}^n$ : with the Euclidean metric a subset of $\mathbb{R}^n$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length, Lower bound, bounded below, bounded set).

[L2]

The sets open in the subspace $[0,1]$ are the traces on $[0,1]$ of the open subsets of $\mathbb{R}$ , and $(-1,2/3)$ and $(1/3,2)$ are open in $\mathbb{R}$ (A subset of a metric space is open in the subspace metric exactly when it is the trace of an open set of the ambient space, and it is compact as a metric space in its own right exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it, 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, Open ball, closed ball and sphere in a metric space, 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).

[L3]

$\operatorname{diam}(A) = \sup\{|u-v| : u,v \in A\}$ for nonempty bounded $A$ , so $|u - v| \le \operatorname{diam}(A)$ for all $u,v \in A$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

[L4]

A Lebesgue number for an open cover of a compact metric space is a real $\delta > 0$ such that every nonempty subset of diameter less than $\delta$ lies in a single member of the cover; one exists (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).

Verification

technique · direct

1.1

$U$ and $V$ are open in $[0,1]$ , being the traces of $(-1,2/3)$ and $(1/3,2)$ , and $U \cup V = [0,1]$ , because a point of $[0,1]$ is either below $2/3$ , and then in $U$ , or at least $2/3 > 1/3$ , and then in $V$ .

L1L2

2.1

Let $A \subseteq [0,1]$ be nonempty with $\operatorname{diam}(A) < 1/3$ , and suppose first that every $y \in A$ satisfies $y < 2/3$ ; then $A \subseteq U$ .

L3step 1.1

3.1

Otherwise some $a \in A$ has $a \ge 2/3$ , and then every $y \in A$ satisfies $y \ge a - |a - y| \ge a - \operatorname{diam}(A) > 2/3 - 1/3 = 1/3$ , so $A \subseteq (1/3,1] = V$ .

L3step 2.1

4.1

In both cases $A$ lies in a single member of $\{U,V\}$ , so $1/3$ is a Lebesgue number: claim 1.

L4step 2.1step 3.1

5.1

For claim 2 let $\delta' > 1/3$ be real and put $A := [1/3, 2/3]$ , a nonempty subset of $[0,1]$ with $|u-v| \le 1/3$ for $u,v \in A$ and with $|2/3 - 1/3| = 1/3$ attained, so $\operatorname{diam}(A) = 1/3 < \delta'$ .

L3step 4.1

6.1

$A \not\subseteq U$ because $2/3 \in A$ and $2/3 \notin [0,2/3)$ , and $A \not\subseteq V$ because $1/3 \in A$ and $1/3 \notin (1/3,1]$ ; so no $\delta' > 1/3$ is a Lebesgue number for this cover, and $1/3$ is the largest one: claim 2.

L4step 5.1∎

Remarks

The number is exactly the overlap. $U \cap V = (1/3, 2/3)$ has length $1/3$ , and that is the Lebesgue number: a set of diameter below the overlap cannot straddle both ends. The example shows that the conclusion of 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 is sharp, the lemma asserting only that some positive $\delta$ exists.

The strict inequality in the definition matters. The set $[1/3,2/3]$ has diameter exactly $1/3$ and lies in neither member, so a Lebesgue number could not be required to work for subsets of diameter at most $\delta$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 118 results over 18 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Lebesgue's number lemma (Wikipedia) (standard reference, not scraped)
Heine-Borel theorem (Wikipedia) (standard reference, not scraped)

The cover of [0,1][0,1][0,1] by (−1,2/3)(-1, 2/3)(−1,2/3) and (1/3,2)(1/3, 2)(1/3,2) has Lebesgue number 1/31/31/3, and no larger one