Proposition

Le $X$ be a set, and suppose \(\{ \mathscr{T}_{\alpha} \}_{\alpha \in A}\) is a collection of topologies on $X$. Show that the intersection $\mathscr{T} = \cap_{\alpha \in A} \mathscr{T}_{\alpha}$ is a topology on $X$.

Solution

• Each $\mathscr{T}_{\alpha}$ contains $\emptyset$ and $X$ because it is a topology. Thus the intersection must contain $\emptyset$ and $X$.
• Let $S \subset \mathscr{T}$. Then $S \subset \mathscr{T}_{\alpha}$ for each $\alpha$. Then \(\bigcup_{U \in S} U \in \mathscr{T}_{\alpha}\) for each $\alpha$. Therefore, \(\bigcup_{U \in S} U \in \mathscr{T}\).
• Let \(U_{\alpha_1}, \cdots, U_{\alpha_n} \subset \mathscr{T}\). Then \(U_{\alpha_1}, \cdots, U_{\alpha_n} \subset \mathscr{T}_{\alpha}\) for each $\alpha$. Then \(U_{\alpha_1} \cap \cdots \cap U_{\alpha_n} \in \mathscr{T}_{\alpha}\) for each $\alpha$. Therefore, \(U_{\alpha_1} \cap \cdots \cap U_{\alpha_n} \in \mathscr{T}\).

Therefore, $\mathscr{T}$ is a topology on $X$.