6. Let ℝω have the uniform topology. Given n, let Bn be the collection of all subsets of the form ∏Ai where Ai=ℝ for i≤n 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 ø∉B, and construct a choice function f:B→∪B such that f(B)∈B for each B∈B. If {Uα} is a countable basis for X and B is locally finite, then the countable set {Vn}={Uα | f−1(Uα) finite} covers X. But now B=f−1(X)=f−1(∪Vn)=∪f−1(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=ℝ for every i≤N, 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 ℝω intersects at most one element of Bn, so Bn is evidently locally finite, hence B is countably locally finite. ◻
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