LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

In a locally compact metric space every point has arbitrarily small compact closed balls, hence a neighbourhood base of compact sets

Statement

Let $(X, d)$ be a locally compact metric space (Locally compact metric space: every point has a compact neighbourhood) and let $x \in X$ . Then there is a real $r_{0} > 0$ such that

$\overset{ˉ}{B} (x, r)$ is a compact subset of $X$ (Open ball, closed ball and sphere in a metric space, Open cover, subcover, compact metric space, and compact subset of a metric space) for every real $r$ with $0 < r < r_{0}$ ; and
the family ${\overset{ˉ}{B} (x, r) : 0 < r < r_{0}}$ is a neighbourhood base at $x$ (Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open) consisting of compact sets: every neighbourhood of $x$ contains one of these closed balls.

Note that the closed ball itself is compact, not merely its closure; a closed ball is already closed (Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed). No choice principle is used.

Facts & Assumptions

Given: A locally compact metric space $(X, d)$ and a point $x \in X$ ; balls $B (x, r)$ and $\overset{ˉ}{B} (x, r)$ as in Open ball, closed ball and sphere in a metric space.

[A1]

Local compactness at $x$ : there are a compact subset $K \subseteq X$ and a real $r_{0} > 0$ with $B (x, r_{0}) \subseteq K$ (Locally compact metric space: every point has a compact neighbourhood).

[L1]

$\overset{ˉ}{B} (x, r)$ is closed in $(X, d)$ for every real $r > 0$ , and a set is closed exactly when its complement is open (Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed, 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).

[L2]

A subset $A \subseteq X$ is compact exactly when the metric subspace $(A, d_{A})$ is a compact metric space, $d_{A}$ being the restriction of $d$ ; and for $A \subseteq K \subseteq X$ the metric $A$ inherits from $(K, d_{K})$ is $d_{A}$ , both being the restriction of $d$ to $A \times A$ (Open cover, subcover, compact metric space, and compact subset of a metric space, Isometry, isometric embedding, and the subspace metric on a subset).

[L3]

A subset of a metric subspace is open in that subspace exactly when it is the trace on it of a set open in the ambient space (A subset of a metric space is open in the subspace metric exactly when it is the trace of an open set of the ambient space, and it is compact as a metric space in its own right exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it, claim 1).

[L4]

A closed subset of a compact metric space is a compact subset of it (A closed subset of a compact metric space is compact).

[L5]

$B (x, s) \subseteq \overset{ˉ}{B} (x, s)$ , and $0 < s \leq t$ gives $B (x, s) \subseteq B (x, t)$ and $\overset{ˉ}{B} (x, s) \subseteq \overset{ˉ}{B} (x, t)$ ; moreover $\overset{ˉ}{B} (x, s) \subseteq B (x, t)$ whenever $0 < s < t$ , since $d (x, y) \leq s < t$ (Open ball, closed ball and sphere in a metric space).

[L6]

A set $N$ is a neighbourhood of $x$ exactly when there is a real $s > 0$ with $B (x, s) \subseteq N$ , the balls around $x$ being a neighbourhood base there (Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open, 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).

[L7]

The minimum of a two-element set of reals exists and is one of the two elements (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

Proof

technique · direct

1.1

Fix $K$ and $r_{0}$ as in [A1], so that $K$ is compact and $B (x, r_{0}) \subseteq K$ .

A1choose

1.2

Let $r$ be a real with $0 < r < r_{0}$ .

given

1.3

For claim 2, let $N$ be a neighbourhood of $x$ and fix a real $s > 0$ with $B (x, s) \subseteq N$ .

L6choose

2.1

$\overset{ˉ}{B} (x, r) \subseteq B (x, r_{0}) \subseteq K$ , the first inclusion because $r < r_{0}$ .

step 1.1step 1.2L5

2.2

$X ∖ \overset{ˉ}{B} (x, r)$ is open in $(X, d)$ , since $\overset{ˉ}{B} (x, r)$ is closed.

step 1.2L1

2.3

Put $r : = \min {s, r_{0}} / 2$ , a real with $0 < r$ , $r < r_{0}$ and $r < s$ , since $\min {s, r_{0}}$ is one of $s$ and $r_{0}$ and is at most each of them.

step 1.3step 1.1L7

3.1

$K ∖ \overset{ˉ}{B} (x, r) = (X ∖ \overset{ˉ}{B} (x, r)) \cap K$ is open in the metric subspace $(K, d_{K})$ , being the trace on $K$ of a set open in $X$ ; hence $\overset{ˉ}{B} (x, r) = \overset{ˉ}{B} (x, r) \cap K$ is closed in $(K, d_{K})$ .

step 2.1step 2.2L2L3

4.1

$(K, d_{K})$ is a compact metric space by step 1.1, so its closed subset $\overset{ˉ}{B} (x, r)$ is a compact subset of it, that is the metric subspace of $(K, d_{K})$ on $\overset{ˉ}{B} (x, r)$ is a compact metric space.

step 1.1step 3.1L2L4

5.1

That metric subspace is $(\overset{ˉ}{B} (x, r), d_{\overset{ˉ}{B} (x, r)})$ , the metric being the restriction of $d$ either way, so $\overset{ˉ}{B} (x, r)$ is a compact subset of $X$ ; this is claim 1.

step 4.1L2

6.1

By step 5.1 the set $\overset{ˉ}{B} (x, r)$ is compact, and $\overset{ˉ}{B} (x, r) \subseteq B (x, s) \subseteq N$ because $r < s$ ; moreover $\overset{ˉ}{B} (x, r)$ is a neighbourhood of $x$ , since $B (x, r) \subseteq \overset{ˉ}{B} (x, r)$ and $B (x, r)$ is open and contains $x$ .

step 5.1step 1.3step 2.3L5L6

7.1

Steps 5.1 and 6.1 give claims 1 and 2: every $\overset{ˉ}{B} (x, r)$ with $0 < r < r_{0}$ is a compact neighbourhood of $x$ , and every neighbourhood of $x$ contains one of them.

step 5.1step 6.1∎

Remarks

Where local compactness is spent. Once, at step 1.1, to produce a single compact set with nonempty interior around $x$ . Everything after that is the hereditary behaviour of compactness: a closed subset of a compact space is compact (A closed subset of a compact metric space is compact), and closed balls are closed (Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed).
The bound $r < r_{0}$ is not cosmetic. The closed ball $\overset{ˉ}{B} (x, r_{0})$ itself need not be contained in $K$ , since $B (x, r_{0}) \subseteq K$ says nothing about points at distance exactly $r_{0}$ , so the argument is run strictly below $r_{0}$ . That costs nothing, because arbitrarily small radii are what the neighbourhood base of claim 2 needs.
Compactness of every closed ball is a strictly stronger property. The lemma asserts compactness of the small closed balls at each point, with the threshold depending on the point. In $R^{n}$ every closed ball is compact (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), but that is a feature of $R^{n}$ and not a consequence of local compactness.

Depends on

Used by

If X is a locally compact metric space then the evaluation map is continuous for the compact-open topology Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 122 results over 19 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Locally compact space (Wikipedia) (standard reference, not scraped)
J. Munkres, Topology, 2nd ed., §29 (standard reference, not scraped)