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A countable coordinate-ball cover has a countable locally finite shrinking
Statement
Let be a countable cover of a smooth manifold by coordinate balls with compact closures. Then there are countable families of open sets and coordinate balls such that , each for some index , and the family is locally finite.
Facts & Assumptions
Given: 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.
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]. Fix , choose some containing , and set By step 2.1, this is an open neighbourhood of . Apply [L2] to choose an open with , and then [L1] to choose a coordinate ball with and compact closure. Thus . Applying [L2] inside , choose an open set with Compactness of gives finitely many such pairs covering .
Collect the finitely many pairs from each into sequences and . They are countable, and they cover because the annuli cover . Given , choose with . If is attached to with , then step 3.1 gives because . Hence the neighbourhood meets only the finitely many families attached to . Thus is locally finite.
Hence and give the required countable locally finite shrinking.
Depends on
- 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
Used by
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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)