PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16

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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.
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Every compact compact-open family is pointwise relatively compact

Statement

Let $X$ be a topological space, let $Y$ be a metric space, and let $K \subseteq C (X, Y)$ be compact in the compact-open topology. Then $K$ is pointwise relatively compact.

Facts & Assumptions

Given: A topological space $X$ , a metric space $Y$ , and a compact compact-open family $K$ .

[L1]

For compact $K \subseteq X$ and open $V \subseteq Y$ , $S (K, V) = {f : f [K] \subseteq V}$ is compact-open subbasic (The compact-open topology on $C (X, Y)$ for arbitrary topological spaces).

[L2]

Pointwise relative compactness asks that $\overline{K (x)}$ be compact for every $x$ (Equicontinuity on a topological domain and pointwise relative compactness).

[L3]

Continuous images of compact spaces are compact (A continuous image of a compact space is compact; a continuous real-valued map on a nonempty compact space attains a maximum and a minimum; and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism).

[L4]

Compact subsets of Hausdorff spaces are closed (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).

[L5]

Every metric space is Hausdorff (Distinct points of a metric space have disjoint balls around them).

Proof

technique · direct

1.1L1L2

If $X = \emptyset$ , [L2] is vacuous. Otherwise fix $x \in X$ . For open $V \subseteq Y$ , evaluation at $x$ pulls $V$ back to $S ({x}, V)$ , which is open by [L1]; hence evaluation is continuous.

2.1L3L4L5step 1.1

By [L3], its image $K (x)$ is compact. By [L4] and [L5], $K (x)$ is closed.

3.1L2step 2.1∎

Therefore $\overline{K (x)} = K (x)$ is compact. Since $x$ was arbitrary, [L2] proves pointwise relative compactness.

Depends on

Dependency tree · 2 levels

Sources

Topology, second edition, Theorem 47.1