Find a point-finite open covering \(\mathcal{A}\) of \(\mathbb{R}\) that is not locally finite. (The collection \(\mathcal{A}\) is point-finite if each point of \(\mathbb{R}\) lies in only finitely many elements of \(\mathcal{A}\).)
Solution
For \(n\ge2\), put \[ U_n=\left(\frac1{n+1},\frac1{n-1}\right), \] and let \[ \mathcal A=\{(-\infty,1),(0,\infty)\}\cup\{U_n:n\ge2\}. \] This is an open covering of \(\mathbb R\): the first two sets already cover \(\mathbb R\).
It is point-finite. If \(x\le0\), then \(x\) lies in none of the \(U_n\). If \(x>0\), choose \(N\) with \(1/(N-1)<x\). Then for every \(n\ge N\), \[ U_n\subset\left(0,\frac1{N-1}\right), \] so \(x\notin U_n\). Hence each point belongs to only finitely many of the \(U_n\), and therefore to only finitely many members of \(\mathcal A\).
However, \(\mathcal A\) is not locally finite at \(0\). Every neighborhood \((-\varepsilon,\varepsilon)\) of \(0\) meets \(U_n\) for all sufficiently large \(n\), since \(1/n\to0\). Thus every neighborhood of \(0\) meets infinitely many members of \(\mathcal A\).