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

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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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$R^{n}$ is closed and unbounded and is not compact for $n \geq 1$

Statement refuted

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

For $n \geq 1$ , $R^{n}$ is closed and unbounded, hence it is not compact.

Facts & Assumptions

Given: $n \geq 1$ , Euclidean space $R^{n}$ , and a standard unit vector $e_{0}$ .

[A1]

Every closed 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 standard vector $e_{0}$ 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]

For every real radius there is a natural number larger than it (Every complete ordered field is Archimedean).

[L4]

The empty set is open, so the whole space is closed; a metric subset is bounded exactly when it lies in some ball about some centre (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison, Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

Counterexample

technique · direct

1.1

The whole space $R^{n}$ is closed, since its complement is empty and the empty set is open.

1.2

It is unbounded. Indeed, for an arbitrary centre $c \in R^{n}$ and radius $r > 0$ , choose a natural $k > r + ∥ c ∥_{2}$ by [L3]. The reverse triangle inequality gives $∥ k e_{0} - c ∥_{2} \geq k - ∥ c ∥_{2} > r,$ so $R^{n}$ is contained in no ball.

L2L3L4choose

2.1

By [L1], the closed unbounded space $R^{n}$ is not compact. Hence it refutes [A1].

A1L1step 1.1step 1.2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

43 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

Heine-Borel theorem (standard reference, not scraped)

RnRn is closed and unbounded and is not compact for n≥1n≥1