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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: every subspace of a locally compact space is locally compact

Statement

False claim: local compactness (Locally compact topological space: every point has a compact neighbourhood; and what this says in a metric space) is a hereditary property (Hereditary, open-hereditary and closed-hereditary properties of topological spaces): every subspace (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) of a locally compact space is locally compact.

Where the claim comes from, and what is actually true. Metrizability is hereditary, and so are several other properties of the same shape, so the expectation is natural. What is true for local compactness is heredity along open subspaces and along closed subspaces of a locally compact Hausdorff space; an arbitrary subspace need not inherit it. The witness below is the rationals inside the real line, which is neither open nor closed in it.

Facts & Assumptions

Given: The real line R\mathbb{R} with its usual topology, the metric dR(s,t)=std_{\mathbb{R}}(s,t) = |s-t|, the subset QR\mathbb{Q} \subseteq \mathbb{R} of rationals with the subspace topology, and the bounded open intervals (c,d)(c,d).

[A1]

The false claim: every subspace of a locally compact space is locally compact.

[L4]

A subset AA of a space is a compact subset when the subspace it carries is compact, and for ASXA \subseteq S \subseteq X the topology AA inherits from SS is the one it inherits from XX, so compactness of AA does not depend on which of the two it is read in (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, Hereditary, open-hereditary and closed-hereditary properties of topological spaces).

Refutation

technique · contradiction
1.1

Suppose the claim [A1] holds, so that every subspace of a locally compact space is locally compact.

A1assume-contra
1.2

R\mathbb{R} with its usual topology is locally compact by [L2].

L1L2
2.1

By [A1] and step 1.2 the subspace Q\mathbb{Q} would be locally compact, so the point 00 of Q\mathbb{Q} would have a compact neighbourhood KK in Q\mathbb{Q}: a set KQK \subseteq \mathbb{Q}, compact as a subspace, containing a set open in Q\mathbb{Q} that contains 00. By [L1] and the definition of the subspace topology that open set contains (ε,ε)Q(-\varepsilon, \varepsilon) \cap \mathbb{Q} for some real ε>0\varepsilon > 0, so (ε,ε)QK(-\varepsilon, \varepsilon) \cap \mathbb{Q} \subseteq K.

A1L1L4step 1.2
3.1

By [L4] the set KK is a compact subset of R\mathbb{R} as well, and hence closed in R\mathbb{R} and bounded by [L5].

L4L5step 2.1
4.1

By [L3] there is an irrational tt with 0<t<ε0 < t < \varepsilon. Every neighbourhood of tt in R\mathbb{R} contains an interval (c,d)(c,d) with c<t<dc < t < d, which may be shrunk so that 0<c0 < c and d<εd < \varepsilon, and [L3] then puts a rational qq with c<q<dc < q < d in it; that qq lies in (ε,ε)QK(-\varepsilon,\varepsilon) \cap \mathbb{Q} \subseteq K. So every neighbourhood of tt meets KK, and KK being closed, [L5] gives tKQt \in K \subseteq \mathbb{Q} — but tt is irrational. This contradiction refutes the claim [A1].

A1L1L3L5step 2.1step 3.1discharge-contradiction

Remarks

Why Q\mathbb{Q} fails at every point, not just at 00. The argument uses nothing about 00 beyond its being rational: for any qQq \in \mathbb{Q} a compact neighbourhood would have to be a closed subset of R\mathbb{R} containing all rationals near qq, and hence would contain the irrationals near qq as well, which it cannot. So Q\mathbb{Q} is nowhere locally compact, and the witness is not an isolated defect at one point.

What the failure is about. A compact subset of R\mathbb{R} is closed in R\mathbb{R}, and a subset of Q\mathbb{Q} that is closed in R\mathbb{R} has empty interior in R\mathbb{R}; so no compact subset of Q\mathbb{Q} can contain a whole interval's worth of rationals. The two facts pull in opposite directions, and Q\mathbb{Q} is caught between them precisely because it is dense in R\mathbb{R} and is not all of it.

Heredity does hold in two special cases and they are proved rather than assumed: along open subspaces of a locally compact Hausdorff space and along closed subspaces of any locally compact space (In a locally compact Hausdorff space every point has a neighbourhood base of compact sets, every open subspace and every closed subspace is locally compact, every open set around a point contains an open set with compact closure inside it, and every compact set sits inside an open set with compact closure, claim 2). The set Q\mathbb{Q} is neither open nor closed in R\mathbb{R}, so it escapes both.

Depends on

Used by

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Sources