LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.
AI-generated — a genuinely novel statement formulated by AI, with no source for the claim itself.

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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 $\mathcal U$ .

[A1]

Countable choice selects from a countably indexed family of nonempty sets (The Axiom of Countable Choice ( $\mathrm{AC}_\omega$ )).

[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 $\sigma$ -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 $U\in\mathcal U$ covers $X$ by [L1]; by Lindelöfness take a sequence $V_0,V_1,\ldots$ covering $X$ .

L1F1construct

2.1

By [A1], choose $U_n\in\mathcal U$ with $\overline{V_n}\subseteq U_n$ for each $n$ .

A1step 1.1choose

3.1

Put $W_n:=U_n\setminus\bigcup_{i<n}\overline{V_i}$ . 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 $W_0,\ldots,W_k$ .

step 3.1

5.1

Thus $\{W_n\}$ is a locally finite open refinement of $\mathcal U$ , 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

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 74 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

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)