Consider \(\mathbb{R}^\omega\) in the uniform topology. Given \(n\), let \(\mathcal{B}_n\) be the collection of all subsets of \(\mathbb{R}^\omega\) of the form \(\prod A_i\), where \(A_i = \mathbb{R}\) for \(i \leq n\) and \(A_i\) equals either \(\ts{0}\) or \(\ts{1}\) otherwise. Show that the collection \(\mathcal{B} = \bigcup \mathcal{B}_n\) is countably locally finite, but neither countable nor locally finite.
Solution
Give \(\mathbb R^\omega\) the uniform metric \[ \bar\rho(x,y)=\sup_i\min\{|x_i-y_i|,1\}. \] Fix \(n\). A member of \(\mathcal B_n\) is determined by a binary tail \[ (\epsilon_{n+1},\epsilon_{n+2},\ldots)\in\{0,1\}^{\{i>n\}}, \] and has the form \[ B_\epsilon=\mathbb R^n\times\prod_{i>n}\{\epsilon_i\}. \]
We first show that \(\mathcal B_n\) is locally finite. Fix \(x\in\mathbb R^\omega\) and use the ball \(B_{\bar\rho}(x,1/3)\). If two members \(B_\epsilon,B_\eta\in\mathcal B_n\) both met this ball, choose \(y\in B_\epsilon\) and \(z\in B_\eta\) in the ball. For every \(i>n\), \[ |\epsilon_i-x_i|<1/3, \qquad |\eta_i-x_i|<1/3. \] Since \(\epsilon_i,\eta_i\in\{0,1\}\), the intervals of radius \(1/3\) around \(0\) and \(1\) are disjoint, so \(\epsilon_i=\eta_i\) for every \(i>n\). Thus \(B_\epsilon=B_\eta\). Hence this ball meets at most one member of \(\mathcal B_n\), so \(\mathcal B_n\) is locally finite. Therefore \[ \mathcal B=\bigcup_{n\ge1}\mathcal B_n \] is countably locally finite.
The collection is not countable: already \(\mathcal B_1\) is in bijection with the set of all binary sequences indexed by \(i>1\), hence has cardinality \(2^{\aleph_0}\).
It is not locally finite. Let \(0=(0,0,\ldots)\). For each \(n\), the set \[ C_n=\mathbb R^n\times\prod_{i>n}\{0\} \] belongs to \(\mathcal B_n\), contains \(0\), and the \(C_n\)’s are pairwise distinct. Thus every neighborhood of \(0\) meets infinitely many members of \(\mathcal B\). Hence \(\mathcal B\) is not locally finite.