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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 , and an open cover .
Countable choice selects from a countably indexed family of nonempty sets (The Axiom of Countable Choice ()).
If with open in a regular space, then some open satisfies (A space is regular if and only if every point has a neighbourhood base of closed neighbourhoods, if and only if open gives an open with ).
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 -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
The family of all open for which for some covers by [L1]; by Lindelöfness take a sequence covering .
By [A1], choose with for each .
Put . Each is open and lies in .
The cover : if is the least index with , then for , since , and hence .
The cover is locally finite: for , the neighbourhood is disjoint from for every , while it can meet only .
Thus is a locally finite open refinement of , and [F1] proves paracompactness.
Depends on
Paracompactness: every open cover has a locally finite open refinement, with no separation axiom built into the word
Countably compact, Lindel"of, sequentially compact, limit point compact and -compact spaces, and relatively compact subsets
A space is regular if and only if every point has a neighbourhood base of closed neighbourhoods, if and only if open gives an open with
The Axiom of Countable Choice ()