Let \(\mathcal{C}\) be a collection of subsets of the set \(X\). Suppose that \(\varnothing\) and \(X\) are in \(\mathcal{C}\), and that finite unions and arbitrary intersections of elements of \(\mathcal{C}\) are in \(\mathcal{C}\). Show that the collection
\[ \mathcal{T} = \ts{X - C \mid C \in \mathcal{C}} \]
is a topology on \(X\).
Solution
We verify the topology axioms for \[ \mathcal T=\{X-C:C\in\mathcal C\}. \] Since \(X,\varnothing\in\mathcal C\), their complements show that \(\varnothing,X\in\mathcal T\).
If \(U_\alpha=X-C_\alpha\in\mathcal T\), then by De Morgan, \[ \bigcup_\alpha U_\alpha =X-\bigcap_\alpha C_\alpha. \] The arbitrary intersection \(\bigcap_\alpha C_\alpha\) belongs to \(\mathcal C\), so the union belongs to \(\mathcal T\).
For finitely many \(U_i=X-C_i\), \[ \bigcap_{i=1}^nU_i=X-\bigcup_{i=1}^nC_i, \] and the finite union on the right belongs to \(\mathcal C\). Hence \(\mathcal T\) is a topology.