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A compact metric space is second-countable under countable choice

Statement

Assume the Axiom of Countable Choice. If $(X, d)$ is a compact metric space, then its metric topology has an at-most-countable basis; equivalently, $X$ is second-countable.

Facts & Assumptions

Given: The Axiom of Countable Choice and a compact metric space $(X, d)$ .

[A1]

The Axiom of Countable Choice selects one member from each member of a countable family of nonempty sets. The Axiom of Countable Choice ( ${A C}_{ω}$ )

[F1]

Under the Axiom of Countable Choice, every compact metric space has an at-most-countable dense subset. A compact metric space has a countable dense subset, by countable choice

[F2]

A product of two at-most-countable sets is at most countable. A product of two at most countable sets is at most countable

[F3]

For every $r > 0$ there is an integer $n \geq 1$ with $1 / n < r$ . For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$

[F4]

A metric-open set contains a positive-radius ball around each of its points, and $B (a, r) = {x : d (x, a) < r}$ . 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 Open ball, closed ball and sphere in a metric space

[F5]

A topological space is second-countable when it has an at-most-countable basis. Second countability: an at most countable basis for the topology

Proof

technique · direct

1.1A1F1F2construct

By [F1], choose an at-most-countable dense set $D \subseteq X$ . If $X = \emptyset$ , the empty family is a countable basis. Otherwise set $U : = {B (a, 1 / n) : a \in D, n \in N, n \geq 1}$ . By [F2], the index set $D \times {n \in N : n \geq 1}$ and therefore $U$ are at most countable.

2.1F1F3F4step 1.1choose

To check that $U$ is a basis, let $O$ be open and $x \in O$ . By [F4], choose $r > 0$ with $B (x, r) \subseteq O$ . Apply [F3] to $r / 2$ to choose $n \geq 1$ with $2 / n < r$ . Density of $D$ gives $a \in D$ with $d (x, a) < 1 / n$ . Then $x \in B (a, 1 / n)$ , and every $y \in B (a, 1 / n)$ satisfies $d (x, y) \leq d (x, a) + d (a, y) < 2 / n < r$ . Hence $x \in B (a, 1 / n) \subseteq O$ .

3.1step 2.1F5∎

By [F5], the at-most-countable basis $U$ makes $X$ second-countable.

Depends on

Used by

Kantorovich–Rubinstein duality on compact metric spaces under countable choice Theorem

Dependency tree · two levels

37 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

Dmitriy Panchenko, Lecture Notes on Probability Theory, §4.4 (standard reference, not scraped)
Filippo Santambrogio, Introduction to Optimal Transport Theory, §§1–2 (standard reference, not scraped)