False statementConstruction: AI-adaptedVerification: AI-generatedprecheck 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 $R$ with its usual topology, the metric $d_{R} (s, t) = ∣ s - t ∣$ , the subset $Q \subseteq R$ of rationals with the subspace topology, and the bounded open intervals $(c, d)$ .

[A1]

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

[L1]

$R$ with its usual topology is metrizable, its open sets being exactly the sets $U$ such that every $x \in U$ has $(x - r, x + r) \subseteq U$ for some real $r > 0$ (The absolute value makes $R$ a metric space: $d (x, y) = ∣ x - y ∣$ is a metric, its open balls are the intervals $(x - r, x + r)$ , and it is unbounded, claim 3; The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

[L2]

$R$ is locally compact: for $p \in R$ the set $[p - 1, p + 1]$ is closed and bounded, hence a compact subset of the metric space $R$ (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, claim 3) and so a compact subset of the topological space $R$ (For a metric space with its metric topology, compactness in the topological sense is compactness in the metric sense, and the two notions of compact subset coincide, claim 2); and it contains the open $(p - 1, p + 1) ∋ p$ , so it is a neighbourhood of $p$ (Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open, Locally compact topological space: every point has a compact neighbourhood; and what this says in a metric space).

[L3]

For reals $c < d$ there is a rational strictly between them (ℚ is dense in every Archimedean ordered field), and there is also an irrational strictly between them, the irrationals being dense in $R$ (Both $Q$ and $R ∖ Q$ are dense in $R$ , and every nonempty open subset of $R$ is uncountable, claim 2; Interior, closure, boundary, exterior, derived set and isolated point in a topological space).

[L4]

A subset $A$ of a space is a compact subset when the subspace it carries is compact, and for $A \subseteq S \subseteq X$ the topology $A$ inherits from $S$ is the one it inherits from $X$ , so compactness of $A$ 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).

[L5]

A compact subset of the metric space $R$ is closed in $R$ and bounded (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, claim 3), and a closed subset of $R$ contains every point of $R$ all of whose neighbourhoods meet it (A point lies in the closure of $A$ iff every basic neighbourhood of it meets $A$ ; the closure is the smallest closed superset and equals $A$ together with its derived set, claim 1; Interior, closure, boundary, exterior, derived set and isolated point in a topological space).

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

$R$ with its usual topology is locally compact by [L2].

L1L2

2.1

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

A1L1L4step 1.2

3.1

By [L4] the set $K$ is a compact subset of $R$ as well, and hence closed in $R$ and bounded by [L5].

L4L5step 2.1

4.1

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

A1L1L3L5step 2.1step 3.1discharge-contradiction∎

Remarks

Why $Q$ fails at every point, not just at $0$ . The argument uses nothing about $0$ beyond its being rational: for any $q \in Q$ a compact neighbourhood would have to be a closed subset of $R$ containing all rationals near $q$ , and hence would contain the irrationals near $q$ as well, which it cannot. So $Q$ is nowhere locally compact, and the witness is not an isolated defect at one point.

What the failure is about. A compact subset of $R$ is closed in $R$ , and a subset of $Q$ that is closed in $R$ has empty interior in $R$ ; so no compact subset of $Q$ can contain a whole interval's worth of rationals. The two facts pull in opposite directions, and $Q$ is caught between them precisely because it is dense in $R$ 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 $Q$ is neither open nor closed in $R$ , so it escapes both.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

85 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Locally compact space (Wikipedia) (standard reference, not scraped)
Heine-Borel theorem (Wikipedia) (standard reference, not scraped)
I. Khatchatourian, Compactifications (MAT327 notes) (standard reference, not scraped)