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

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Under countable choice, every regular Lindelöf space is paracompact

Statement

Assume the Axiom of Countable Choice. Every regular Lindelöf topological space is paracompact.

Facts & Assumptions

Given: Countable choice, a regular Lindelöf space $X$ , and an open cover $U$ .

[A1]

Countable choice selects from a countably indexed family of nonempty sets (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[L1]

If $x \in U$ with $U$ open in a regular space, then some open $V$ satisfies $x \in V \subseteq \overline{V} \subseteq U$ (A space is regular if and only if every point has a neighbourhood base of closed neighbourhoods, if and only if $x \in U$ open gives an open $V$ with $x \in V \subseteq \overline{V} \subseteq U$ ).

[F1]

Lindelöfness gives an at most countable subcover, and paracompactness asks for a locally finite open refinement (Countably compact, Lindel"of, sequentially compact, limit point compact and $σ$ -compact spaces, and relatively compact subsets, Paracompactness: every open cover has a locally finite open refinement, with no separation axiom built into the word).

Proof

technique · constructive

1.1

The family of all open $V$ for which $\overline{V} \subseteq U$ for some $U \in U$ covers $X$ by [L1]; by Lindelöfness take a sequence $V_{0}, V_{1}, \dots$ covering $X$ .

L1F1construct

2.1

By [A1], choose $U_{n} \in U$ with $\overline{V_{n}} \subseteq U_{n}$ for each $n$ .

A1step 1.1choose

3.1

Put $W_{n} : = U_{n} ∖ ⋃_{i < n} \overline{V_{i}}$ . Each $W_{n}$ is open and lies in $U_{n}$ .

step 2.1construct

4.1

The $W_{n}$ cover $X$ : if $n$ is the least index with $x \in U_{n}$ , then $x \notin \overline{V_{i}}$ for $i < n$ , since $\overline{V_{i}} \subseteq U_{i}$ , and hence $x \in W_{n}$ .

step 1.1step 2.1step 3.1

4.2

The cover is locally finite: for $x \in V_{k}$ , the neighbourhood $V_{k}$ is disjoint from $W_{n}$ for every $n > k$ , while it can meet only $W_{0}, \dots, W_{k}$ .

step 3.1

5.1

Thus ${W_{n}}$ is a locally finite open refinement of $U$ , and [F1] proves paracompactness.

F1step 3.1step 4.1step 4.2discharge-construct∎

Depends on

Used by

Assuming choice, two paracompact lower-limit lines can have a nonparacompact product Counterexample
Assuming choice, refuted: paracompactness is productive False statement
Topological manifolds are metrizable and paracompact Theorem

Dependency tree · two levels

24 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

Dartmouth Point-Set Topology, Lecture 25 (standard reference, not scraped)
Topology 262 notes (California State University, Northridge) (standard reference, not scraped)
R. Gardner, Notes on Munkres Section 41: Paracompactness (East Tennessee State University) (standard reference, not scraped)