6. Let $ℝ^ω$ have the uniform topology. Given $n$, let $\mathcal{B}_n$ be the collection of all subsets of the form $\prod A_i$ where $A_i=ℝ$ for $i≤n$ and $A_i$ equals $\{0\}$ or $\{1\}$ otherwise. Show $\mathcal{B}=∪\mathcal{B}_n$ is countably locally finite, but is neither countable nor locally finite.
Proof: (5) It suffices to consider when $\mathcal{A}$ 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 $\mathcal{B}$. Assume $ø∉\mathcal{B}$, and construct a choice function $f : \mathcal{B}→∪\mathcal{B}$ such that $f(B)∈B$ for each $B∈\mathcal{B}$. If $\{U_α\}$ is a countable basis for $X$ and $\mathcal{B}$ is locally finite, then the countable set $$\{V_n\}=\{U_α~|~f^{-1}(U_α)\text{ finite}\}$$ covers $X$. But now $$\mathcal{B}=f^{-1}(X)=f^{-1}(∪V_n)=∪f^{-1}(V_n)$$ is a countable union of finite sets, so $\mathcal{B}$ is countable, a contradiction.
(6) It's clear that $\mathcal{B}_n$ is uncountable for any $n$, so $\mathcal{B}$ is not countable. As well, $1^ω$ is contained in every subset of the form $\prod A_i$ where, for some $N$, $A_i=ℝ$ for every $i≤N$, and $A_i=\{1\}$ otherwise, so that $\mathcal{B}$ is not even point-finite, let alone locally finite. But the $1/2$-neighborhood of any element in $ℝ^ω$ intersects at most one element of $\mathcal{B}_n$, so $\mathcal{B}_n$ is evidently locally finite, hence $\mathcal{B}$ is countably locally finite.$~\square$
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