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Friday, December 26, 2014

Countable Local Finiteness (6.39.5-6)

James Munkres Topology, chapter 6.39, exercises 5-6:

MathJax TeX Test Page 5. If X is second-countable, show a collection A of subsets of X is countably locally finite if and only if it is countable.

6. Let ω have the uniform topology. Given n, let Bn be the collection of all subsets of the form Ai where Ai= for in and Ai equals {0} or {1} otherwise. Show B=Bn is countably locally finite, but is neither countable nor locally finite.

Proof: (5) It suffices to consider when A is an uncountable, countably locally finite collection of subsets of X. Since countable unions of countable sets are countable, there exists an uncountable locally finite collection B. Assume øB, and construct a choice function f:BB such that f(B)B for each BB. If {Uα} is a countable basis for X and B is locally finite, then the countable set {Vn}={Uα | f1(Uα) finite} covers X. But now B=f1(X)=f1(Vn)=f1(Vn) is a countable union of finite sets, so B is countable, a contradiction.

(6) It's clear that Bn is uncountable for any n, so B is not countable. As well, 1ω is contained in every subset of the form Ai where, for some N, Ai= for every iN, and Ai={1} otherwise, so that B is not even point-finite, let alone locally finite. But the 1/2-neighborhood of any element in ω intersects at most one element of Bn, so Bn is evidently locally finite, hence B is countably locally finite. 

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