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A countable coordinate-ball cover has a countable locally finite shrinking
Statement
Assume countable choice . Let be a countable cover of a smooth manifold by coordinate balls with compact closures. Then there is an at-most-countable index set and families of open sets and coordinate balls such that , each for some index , and the family is locally finite. The index set may be finite or empty.
Facts & Assumptions
Given: Countable choice and a countable cover of by coordinate balls with compact closures.
Coordinate balls form a basis of the underlying topological manifold, and the basis balls supplied there have compact closures (Coordinate balls form a basis of a topological manifold).
In a regular space, if with open, then there is an open set with (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 ).
Compact subsets of Hausdorff spaces are closed, and closed subspaces of compact spaces are compact (In a Hausdorff space a point and a disjoint compact set, and two disjoint compact sets, have disjoint open neighbourhoods; hence every compact subset is closed, and in a compact Hausdorff space the compact subsets are exactly the closed ones, A closed subspace of a compact space is compact, and a finite union of compact subspaces is compact).
Smooth manifolds are Hausdorff and regular.
Countable choice selects one finite covering list for each compact annulus from independently specified nonempty sets of lists (The Axiom of Countable Choice ()).
Proof
Put for . Each is compact, and the interiors of the cover because .
Set . Recursively, compactness of and the open cover give an integer such that . Put and . Then and the interiors of the cover .
Put and for ; each is compact by [L3]. For each , consider all tuples with a coordinate ball, open, and Their -sets cover : for an individual , take any containing and successively use [L2] and [L1] inside the displayed open neighbourhood to construct such and around . This proves existence at each point without a simultaneous point-indexed choice. By compactness, the set of finite ordered lists of these tuples whose -sets cover is nonempty (and contains the empty list when ).
Use [A2] once on the countable family of nonempty finite-list sets from step 3.1. Concatenate the chosen lists into an at-most-countable indexed family ; take when every list is empty. They cover because the annuli cover . Given , choose with . If came from annulus with , then because . Only finitely many tuples came from the remaining finitely many annuli, so is locally finite.
Hence and give the required countable locally finite shrinking.
Depends on
- The Axiom of Countable Choice ($\mathrm{AC}_\omega$)
- Smooth manifolds and their smooth charts
- Coordinate balls form a basis of a topological manifold
- 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 $x \in V \subseteq \overline{V} \subseteq U$
- In a Hausdorff space a point and a disjoint compact set, and two disjoint compact sets, have disjoint open neighbourhoods; hence every compact subset is closed, and in a compact Hausdorff space the compact subsets are exactly the closed ones
- A closed subspace of a compact space is compact, and a finite union of compact subspaces is compact
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Sources
- John M. Lee, Introduction to Smooth Manifolds (standard reference, not scraped)
- Will J. Merry, Differential Geometry (standard reference, not scraped)
- Nigel Hitchin, Differentiable Manifolds (standard reference, not scraped)