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False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-08-05 (gpt-5.6-sol-codex-subscription)
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FALSE: a compact subset of a topological space is closed

Statement

False claim: in every topological space (X,T) (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison), a compact subset (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right) is closed.

Where the claim comes from, and what is actually true. In a Hausdorff space a compact subset is closed, and that is In a Hausdorff space a point and a disjoint compact set, and two disjoint compact sets, have disjoint open neighbourhoods; hence every compact subset is closed, and in a compact Hausdorff space the compact subsets are exactly the closed ones, claim 3. The claim above is that theorem with its hypothesis dropped. The refutation builds its own witness: Sierpinski space, the two-point space with exactly one non-trivial open set (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).

Facts & Assumptions

Given: The two-element set S={a,b} with a≠b, and the family TSier={∅,{b},S}.

[A1]

The false claim: in every topological space a compact subset is closed.

[L1]

TSier is a topology on S, the particular-point topology with particular point b; a subset of S is closed exactly when its complement lies in TSier (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies, Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

Refutation

technique · contradiction
1.1

Suppose the claim [A1] holds, so that in every topological space every compact subset is closed.

A1assume-contra
1.2

(S,TSier) is a topological space by [L1], and its closed sets are S, {a} and ∅, the complements of ∅, {b} and S.

L1
2.1

{b} is a compact subset of S: the subspace it carries is a one-point space, which is compact by [L2].

L2step 1.2
2.2

{b} is not closed in S, since its complement {a} is not a member of TSier.

L1step 1.2
3.1

By [A1] applied to the space of step 1.2 and the compact subset of step 2.1, the set {b} would be closed, which step 2.2 denies. So the claim [A1] is false.

A1step 2.1step 2.2discharge-contradiction∎

Remarks

The witness is as small as a witness can be. Sierpinski space has two points and three open sets, and it fails the Hausdorff condition for the only reason available: the only open set containing a is S, which also contains b (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not). Since every finite space is compact, every subset of it is a compact subset, so the failure is not about compactness being hard to achieve; it is entirely about closedness.

What survives without a separation hypothesis. A compact subset remains compact in any other ambient inducing the same topology on it — in particular, compactness is invariant under homeomorphism — that being the content of the intrinsic definition (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right), and a closed subset of a compact space is still compact (A closed subspace of a compact space is compact, and a finite union of compact subspaces is compact). It is only the converse direction, from compact to closed, that needs the ambient space to separate points.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources