TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16

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Under Choice, equicontinuity and pointwise relative compactness give compact compact-open closure

Statement

Assume the Axiom of Choice. Let $X$ be a topological space, let $Y$ be a metric space, and let $F \subseteq C (X, Y)$ be equicontinuous and pointwise relatively compact. Then the closure of $F$ in the compact-open topology is compact.

Facts & Assumptions

Given: Choice, a topological space $X$ , a metric space $Y$ , and an equicontinuous, pointwise relatively compact family $F \subseteq C (X, Y)$ .

[L1]

Under Choice, the pointwise closure is compact exactly when every coordinate set has compact closure (Under Choice, pointwise closure is compact exactly when every coordinate set has compact closure).

[L2]

The pointwise closure of an equicontinuous family is equicontinuous and consists of continuous maps (The pointwise closure of an equicontinuous family is equicontinuous and consists of continuous maps).

[L3]

The compact-open and pointwise subspace topologies agree on an equicontinuous family (The compact-open and pointwise topologies agree on an equicontinuous family).

[L4]

Compact-open subbasic sets test compact subsets of the domain (The compact-open topology on $C (X, Y)$ for arbitrary topological spaces).

[L5]

Pointwise subbasic sets test one coordinate at a time (The topology of pointwise convergence on $Y^{X}$ , which is the product topology, and its restriction to $C (X, Y)$ ).

Proof

technique · direct

1.1

Let $H$ be the pointwise closure of $F$ in $Y^{X}$ . Pointwise relative compactness and [L1] make $H$ compact in the pointwise topology.

2.1

By [L2], $H \subseteq C (X, Y)$ and $H$ is equicontinuous. By [L3], its compact-open subspace topology equals its pointwise subspace topology, so $H$ is compact in the compact-open topology.

L2L3step 1.1

3.1

The family $F$ is pointwise dense in $H$ , and [L3] makes it compact-open dense there. Also, every pointwise subbasic set from [L5] is the compact-open set $S ({x}, V)$ of [L4], so the compact-open topology is finer and the pointwise-closed set $H$ is compact-open closed. Thus the compact-open closure of $F$ is exactly $H$ , which is compact by step 2.1.

L3L4L5step 2.1∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 55 results over 16 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

Topology, second edition, Theorem 47.1 (standard reference, not scraped)