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Under Choice, pointwise closure is compact exactly when every coordinate set has compact closure

Statement

Assume the Axiom of Choice. Let $X$ be a set, let $Y$ be a metric space, let $F \subseteq Y^{X}$ , and let $H$ be the closure of $F$ in the topology of pointwise convergence. Then $H$ is compact if and only if $\overline{F (x)}$ is compact in $Y$ for every $x \in X$ .

Facts & Assumptions

Given: The Axiom of Choice, a set $X$ , a metric space $Y$ , and $F \subseteq Y^{X}$ with pointwise closure $H$ .

[L1]

Pointwise convergence is the product topology on $Y^{X}$ , and the coordinate maps $π_{x} (f) = f (x)$ are continuous (The topology of pointwise convergence on $Y^{X}$ , which is the product topology, and its restriction to $C (X, Y)$ ).

[L2]

Under Choice, a product of compact topological spaces is compact (Tychonoff's theorem: an arbitrary product of compact spaces is compact in the product topology, assuming the Axiom of Choice).

[L3]

A closed subspace of a compact space is compact (A closed subspace of a compact space is compact, and a finite union of compact subspaces is compact).

[L4]

A compact subset of a Hausdorff space is 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]

A continuous image of a compact space is 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).

[L6]

Pointwise relative compactness means that every coordinate set has compact closure (Equicontinuity on a topological domain and pointwise relative compactness).

[L7]

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

Proof

technique · direct

1.1

Suppose $H$ is compact and fix $x \in X$ . By [L1] and [L5], $π_{x} [H]$ is compact, and by [L4] and [L7] it is closed.

L1L4L5L7

1.2

Conversely suppose $K_{x} : = \overline{F (x)}$ is compact for every $x$ . By [L2], $P : = \prod_{x \in X} K_{x}$ is compact, including when $X = \emptyset$ , when it is a singleton.

1.3

Each $K_{x}$ is closed in the metric space $Y$ by [L4] and [L7], so $P = ⋂_{x \in X} π_{x}^{- 1} [K_{x}]$ is closed in $Y^{X}$ . Since $F \subseteq P$ , its closure $H$ is a closed subset of $P$ .

L1L4L7

2.1

Since $F (x) \subseteq π_{x} [H]$ , its closure is contained in $π_{x} [H]$ ; conversely continuity gives $π_{x} [H] \subseteq \overline{F (x)}$ . Hence $\overline{F (x)} = π_{x} [H]$ is compact.

step 1.1L6

3.1

By [L3], $H$ is compact. Together with steps 1.1--1.2 this proves both directions.

L3step 1.2step 1.3∎

Depends on

Used by

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

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 88 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, Lemma 47.4 (standard reference, not scraped)