False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck 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 $(X, \mathcal{T})$ (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 $a \ne b$ , and the family $\mathcal{T}_{\mathrm{Sier}} = \{\varnothing, \{b\}, S\}$ .

[A1]

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

[L1]

$\mathcal{T}_{\mathrm{Sier}}$ 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 $\mathcal{T}_{\mathrm{Sier}}$ (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).

[L2]

A subset $A$ of a space is a compact subset when the subspace $(A, \mathcal{T}_A)$ is compact, and every space listed as $\{x_0, \dots, x_n\}$ is compact (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right, Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace).

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

$(S, \mathcal{T}_{\mathrm{Sier}})$ is a topological space by [L1], and its closed sets are $S$ , $\{a\}$ and $\varnothing$ , the complements of $\varnothing$ , $\{b\}$ and $S$ .

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 $\mathcal{T}_{\mathrm{Sier}}$ .

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

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Sources

Compact space (Wikipedia) (standard reference, not scraped)
Sierpiński space (Wikipedia) (standard reference, not scraped)
Stacks Project, Section 5.12: Quasi-compact spaces and maps (standard reference, not scraped)