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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$\mathbb{R}^n$ is closed and unbounded and is not compact for $n\ge1$

Statement refuted

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

For $n\ge1$ , $\mathbb R^n$ is closed and unbounded, hence it is not compact.

Facts & Assumptions

Given: $n\ge1$ , Euclidean space $\mathbb R^n$ , 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 $\mathbb{R}^n$ with $n\ge1$ , compactness, closedness and boundedness, pseudocompactness, and attainment of extrema by every continuous real-valued function are equivalent).

[L2]

The 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 $j \ne i$ 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]

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 $\mathbb R^n$ is closed, since its complement is empty and the empty set is open.

1.2

It is unbounded. Indeed, for an arbitrary centre $c\in\mathbb R^n$ and radius $r>0$ , choose a natural $k>r+\lVert c\rVert_2$ by [L3]. The reverse triangle inequality gives $\lVert ke_0-c\rVert_2\ge k-\lVert c\rVert_2>r,$ so $\mathbb R^n$ is contained in no ball.

L2L3L4choose

2.1

By [L1], the closed unbounded space $\mathbb R^n$ 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 · next 3 levels

Direct dependencies and their dependencies through the next three levels: 126 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

Heine-Borel theorem (standard reference, not scraped)

Rn\mathbb{R}^nRn is closed and unbounded and is not compact for n≥1n\ge1n≥1