Give an example to show that if \(X\) is paracompact, it does not follow that for every open covering \(\mathcal{A}\) of \(X\), there is a locally finite subcollection of \(\mathcal{A}\) that covers \(X\).
Solution
Take \(X=\mathbb R\), which is metrizable and hence paracompact, and consider the open cover \[ \mathcal U=\{(-n,n):n\in\mathbb Z_+\}. \] Any subcollection that covers \(\mathbb R\) must contain intervals with arbitrarily large \(n\), so it must be infinite. But every member of \(\mathcal U\) contains \(0\). Hence any infinite subcollection fails even to be point-finite at \(0\), and therefore cannot be locally finite. Thus this open cover has no locally finite subcover, although \(X\) is paracompact.