TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31

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Assuming countable choice, a metrizable space is second countable if and only if it is separable if and only if it is Lindelöf

Statement

Assuming ${A C}_{ω}$ , a metrizable space is second countable iff it is separable iff it is Lindelöf.

Facts & Assumptions

Given: A metric inducing the topology of $X$ (Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not, 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).

[L1]

Second countability implies Lindelöf under countable choice (Assuming countable choice, every second countable space is Lindelöf).

[A1]

Countable choice selects one object from each nonempty family in a sequence (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[L2]

Under countable choice, a countable union of at most countable sets is at most countable (Countable unions of at most countable sets, assuming ${A C}_{ω}$ ).

Proof

technique · direct

1.1

Suppose $D$ is at most countable and dense. If $D = \emptyset$ , then $X = \emptyset$ and the empty family is a basis. Otherwise the family $B = {B (d, 1 / n) : d \in D, n \geq 1}$ is at most countable. It is a basis: if $x \in U$ with $U$ open, choose $ε > 0$ with $B (x, ε) \subseteq U$ , then choose $n$ with $2 / n < ε$ and $d \in D \cap B (x, 1 / n)$ . Now $x \in B (d, 1 / n) \subseteq B (x, 2 / n) \subseteq U$ . Thus separability implies second countability.

given

1.2

[L1] gives second countable implies Lindelöf.

1.3

Suppose $X$ is Lindelöf. For each $n \geq 1$ , the radius- $1 / n$ balls cover $X$ . Using [A1], choose an at most countable set $D_{n}$ of centres whose radius- $1 / n$ balls cover $X$ . Then $D = ⋃_{n \geq 1} D_{n}$ is at most countable by [L2]. It is dense: for $x \in U$ open, choose $ε > 0$ with $B (x, ε) \subseteq U$ and $n$ with $1 / n < ε$ ; some $d \in D_{n}$ has $x \in B (d, 1 / n)$ , so $d \in B (x, ε) \subseteq U$ . Thus Lindelöf implies separable.

A1L2given

2.1

The three implications prove the equivalence.

step 1.1step 1.2step 1.3∎

Depends on

Used by

Dependency tree · two levels

44 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

UCR General Topology Notes (standard reference, not scraped)