Let \(\mcs, \mct\) be topologies on a set \(X\). Show that \(\mcs \cap \mct\) is a topology on \(X\).
Give an example to show that \(\mcs \cup \mct\) need not be a topology.
Solution
1. The collection \(\mcs\cap\mct\) contains \(\emptyset\) and \(X\).
Since \(\mcs\) and \(\mct\) are topologies on \(X\), \[ \emptyset,X\in\mcs \qquad\text{and}\qquad \emptyset,X\in\mct. \] Hence \[ \emptyset,X\in\mcs\cap\mct. \]
2. The collection \(\mcs\cap\mct\) is closed under arbitrary unions.
Let \(\{U_i\}_{i\in I}\) be any family of sets in \(\mcs\cap\mct\). Then every \(U_i\) belongs to both \(\mcs\) and \(\mct\). Because each is a topology, \[ \bigcup_{i\in I}U_i\in\mcs \qquad\text{and}\qquad \bigcup_{i\in I}U_i\in\mct. \] Therefore \[ \bigcup_{i\in I}U_i\in\mcs\cap\mct. \]
3. The collection \(\mcs\cap\mct\) is closed under finite intersections.
If \(U_1,\dots,U_r\in\mcs\cap\mct\), then all of them belong to each of \(\mcs\) and \(\mct\). Thus \[ U_1\cap\cdots\cap U_r\in\mcs \qquad\text{and}\qquad U_1\cap\cdots\cap U_r\in\mct, \] so \[ U_1\cap\cdots\cap U_r\in\mcs\cap\mct. \]
4. Hence \(\mcs\cap\mct\) is a topology on \(X\).
Steps <1>1–<1>3 are exactly the topology axioms.
5. The union of two topologies need not be a topology.
Let \[ X=\{a,b,c\}, \] and define \[ \mcs=\{\emptyset,X,\{a\}\}, \qquad \mct=\{\emptyset,X,\{b\}\}. \] Each collection is a topology: apart from \(\emptyset\) and \(X\), it has only one nontrivial open set, so it is plainly closed under arbitrary unions and finite intersections.
However, \[ \{a\},\{b\}\in\mcs\cup\mct, \] while \[ \{a\}\cup\{b\}=\{a,b\}\notin\mcs\cup\mct. \] Thus \(\mcs\cup\mct\) is not closed under unions and is not a topology.