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

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

Statement

Let $X$ be a topological space, let $Y$ be a metric space, and let $F \subseteq C (X, Y)$ be equicontinuous. The closure $H$ of $F$ in $Y^{X}$ with the topology of pointwise convergence is equicontinuous, and every $h \in H$ is continuous.

Facts & Assumptions

Given: A topological space $X$ , a metric space $Y$ , and an equicontinuous family $F \subseteq C (X, Y)$ with pointwise closure $H$ .

[L1]

Equicontinuity supplies, for fixed $x$ and tolerance, one neighbourhood of $x$ that works for every member of the family (Equicontinuity on a topological domain and pointwise relative compactness).

[L2]

A basic pointwise neighbourhood controls finitely many coordinate values (The topology of pointwise convergence on $Y^{X}$ , which is the product topology, and its restriction to $C (X, Y)$ ).

[L3]

Coordinate inverse images of open sets are subbasic open in a product topology (The product set $\prod_{i \in I} X_{i}$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space).

Proof

technique · direct

1.1L1

Fix $x \in X$ and $ε > 0$ . By [L1], choose a neighbourhood $U$ of $x$ such that $d (f (y), f (x)) < ε / 3$ for every $f \in F$ and $y \in U$ .

1.2L2L3

Fix $h \in H$ and $y \in U$ . The pointwise neighbourhood of $h$ requiring both $d (g (x), h (x)) < ε / 3$ and $d (g (y), h (y)) < ε / 3$ meets $F$ ; choose $f$ in the intersection.

2.1step 1.1step 1.2

The triangle inequality and steps 1.1--1.2 give $d (h (y), h (x)) < ε$ . The neighbourhood $U$ did not depend on $h$ , so it proves equicontinuity of all of $H$ .

3.1step 2.1∎

For each fixed $h \in H$ , the estimate in step 2.1 is the neighbourhood criterion for continuity at every $x$ . Hence every member of $H$ is continuous.

Depends on

Used by

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

Dependency tree · two levels

17 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

Topology, second edition, Lemma 47.3 (standard reference, not scraped)