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 $a \neq b$ , and the family $T_{S i e r} = {\emptyset, {b}, S}$ .

[A1]

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

[L1]

$T_{S i e r}$ 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 $T_{S i e r}$ (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 $(A, T_{A})$ 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

1.1A1assume-contra

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

1.2L1

$(S, T_{S i e r})$ is a topological space by [L1], and its closed sets are $S$ , ${a}$ and $\emptyset$ , the complements of $\emptyset$ , ${b}$ and $S$ .

2.1L2step 1.2

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

2.2L1step 1.2

${b}$ is not closed in $S$ , since its complement ${a}$ is not a member of $T_{S i e r}$ .

3.1A1step 2.1step 2.2discharge-contradiction∎

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.

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.

FALSE: a compact subset of a topological space is closed

Statement

Facts & Assumptions

Refutation

Remarks

Depends on

Dependency tree · 2 levels

Sources