CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31

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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.
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The open unit ball in $\mathbb{R}^n$ is bounded and not compact

Statement refuted

Refuted claim: every bounded subset of $\mathbb R^n$ is compact.

For $n\ge1$ , the open unit ball $B_2(0,1)$ is bounded but not compact.

Facts & Assumptions

Given: $n\ge1$ , the open unit ball $B_2(0,1)$ , and a standard unit vector $e_0$ .

[A1]

Every bounded Euclidean subset is compact.

[L1]

Euclidean compactness is equivalent to closedness and boundedness (For a nonempty subset of $\mathbb{R}^n$ with $n\ge1$ , compactness, closedness and boundedness, pseudocompactness, and attainment of extrema by every continuous real-valued function are equivalent).

[L2]

The unit vector $e_0$ exists and has Euclidean norm $1$ (The standard list $e : n \to F^{n}$ with $e_i(i) = 1_F$ and $e_i(j) = 0_F$ for $j \ne i$ is an ordered basis of $F^{n}$ ; hence $\dim_F F^{n} = n$ , and $F^{0}$ is the zero space with basis $\varnothing$ and dimension $0$ ).

[L3]

The open ball $B_2(0,1)$ consists of the points of Euclidean distance less than $1$ from $0$ , and metric balls form neighbourhoods in the metric topology (Open ball, closed ball and sphere in a metric space, 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).

Counterexample

technique · direct

1.1

The open unit ball is bounded, since every one of its points has distance less than $1$ , hence less than $2$ , from $0$ .

1.2

It is not closed: $e_0\notin B_2(0,1)$ , while for every $r>0$ the point $(1-\varepsilon)e_0$ , with $0<\varepsilon<\min(r,1)$ , lies in both $B_2(0,1)$ and $B_2(e_0,r)$ . Thus every neighbourhood of $e_0$ meets the open unit ball.

L2L3

2.1

By [L1], the bounded nonclosed set $B_2(0,1)$ is not compact. It therefore refutes [A1].

A1L1step 1.1step 1.2∎

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: 110 results over 23 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

Open ball (standard reference, not scraped)
Heine-Borel theorem (Wikipedia) (standard reference, not scraped)

The open unit ball in Rn\mathbb{R}^nRn is bounded and not compact