Show that if \(\mathcal{A}\) is a basis for a topology on \(X\), then the topology generated by \(\mathcal{A}\) equals the intersection of all topologies on \(X\) that contain \(\mathcal{A}\). Prove the same if \(\mathcal{A}\) is a subbasis.
Solution
Let \(\mathcal T\) be the topology generated by a basis \(\mathcal A\), and let \[ \mathcal S=\bigcap\{\mathcal U:\mathcal U\text{ is a topology on }X\text{ and }\mathcal A\subseteq\mathcal U\}. \] Since \(\mathcal T\) itself is one of the topologies in the intersection, \(\mathcal S\subseteq\mathcal T\). Conversely, if \(U\in\mathcal T\), then \(U\) is a union of elements of \(\mathcal A\). Every topology containing \(\mathcal A\) therefore contains \(U\), hence \(U\in\mathcal S\). So \(\mathcal T=\mathcal S\).
If \(\mathcal A\) is a subbasis, the same first inclusion holds. The basis generated by \(\mathcal A\) consists of finite intersections of elements of \(\mathcal A\), and every topology containing \(\mathcal A\) contains all such finite intersections and all their unions. Therefore every open set in the topology generated by the subbasis lies in every topology containing \(\mathcal A\). Again the two collections are equal.