(May 2016) Let \(X\) be the set of real numbers, let \(\mathcal{T}_E\) be the Euclidean topology on \(X\), and let \(\mathcal{T}_0\) be the excluded point topology (that is, \(\mathcal{T}_0=\{U\subset X\mid 0\notin U\}\cup\{X\}\)). For each of the following topological spaces, determine whether or not the space is compact:
The set \(X\) with the topology \(\mathcal{T}_E\cap\mathcal{T}_0\).
The set \(X\) with the topology generated by the subbasis \(\mathcal{T}_E\cup\mathcal{T}_0\).
Solution
An open set belongs to \(\mathcal T_E\cap\mathcal T_0\) precisely when it is Euclidean-open and either does not contain \(0\), or is all of \(X\). Hence the only open set in this topology that contains \(0\) is \(X\) itself. Every open cover of \(X\) must therefore contain \(X\), and so has the one-element subcover \(\{X\}\). Thus \[ (X,\mathcal T_E\cap\mathcal T_0) \] is compact.
Let \(\mathcal T\) be the topology generated by \(\mathcal T_E\cup\mathcal T_0\). Every singleton \(\{x\}\) with \(x\ne0\) belongs to \(\mathcal T_0\), hence is \(\mathcal T\)-open. Also \((-1,1)\) is \(\mathcal T_E\)-open. Therefore \[ \mathcal U=\{(-1,1)\}\cup\bigl\{\{x\}:|x|\ge1\bigr\} \] is an open cover of \(\mathbb R\). No finite subcollection covers all points with \(|x|\ge1\). Hence \((X,\mathcal T)\) is not compact.