Show that the collection \(\mathcal{T}_c\) given in Example 4 of §12 is a topology on the set \(X\). Is the collection
\[ \mathcal{T}_\infty = \ts{U \mid X - U \text{ is infinite or empty or all of } X} \]
a topology on \(X\)?
Solution
Let \[ \mathcal T_c=\{\varnothing\}\cup\{U\subseteq X: X-U\text{ is countable}\}. \] The complement of \(X\) is empty, so \(\varnothing,X\in\mathcal T_c\). If \(\{U_\alpha\}\) is a family in \(\mathcal T_c\) and all \(U_\alpha\) are empty, its union is \(\varnothing\in\mathcal T_c\). Otherwise choose one nonempty \(U_\beta\). Then \[ X-\bigcup_\alpha U_\alpha=\bigcap_\alpha(X-U_\alpha)\subseteq X-U_\beta, \] so the complement of the union is countable. For finitely many nonempty \(U_1,\dots,U_n\), \[ X-\bigcap_{i=1}^nU_i=\bigcup_{i=1}^n(X-U_i), \] a finite union of countable sets. Hence \(\mathcal T_c\) is a topology.
The analogous collection \(\mathcal T_\infty\) is not a topology in general. Take \(X=\mathbb Z_+\), \[ U=\{2,4,6,\dots\},\qquad V=\{3,5,7,\dots\}. \] Both complements are infinite, so \(U,V\in\mathcal T_\infty\), but \[ U\cup V=X-\{1\}, \] whose complement is the nonempty finite set \(\{1\}\). Thus \(U\cup V\notin\mathcal T_\infty\).