CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31

How statement and proof provenance work

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 open unit ball in $R^{n}$ is bounded and not compact

Statement refuted

Refuted claim: every bounded subset of $R^{n}$ is compact.

For $n \geq 1$ , the open unit ball $B_{2} (0, 1)$ is bounded but not compact.

Facts & Assumptions

Given: $n \geq 1$ , 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 $R^{n}$ with $n \geq 1$ , 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 $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ 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

1.1L3

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

1.2L2L3

It is not closed: $e_{0} \notin B_{2} (0, 1)$ , while for every $r > 0$ the point $(1 - ε) e_{0}$ , with $0 < ε < \min (r, 1)$ , lies in both $B_{2} (0, 1)$ and $B_{2} (e_{0}, r)$ . Thus every neighbourhood of $e_{0}$ meets the open unit ball.

2.1A1L1step 1.1step 1.2∎

By [L1], the bounded nonclosed set $B_{2} (0, 1)$ is not compact. It therefore refutes [A1].

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

33 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

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

The open unit ball in RnRn is bounded and not compact