Proposition

Show that the disjoint union topology is indeed a topology.

Solution

Let \((X_{\alpha})_{\alpha \in A}\) be an indexed family of nonempty topological spaces. Let \(\mathscr{T}_{\alpha}\) be the topology on $X_{\alpha}$ for each $\alpha \in A$. Let $\mathscr{T}$ denote the disjoint union topology on $\sqcup_{\alpha \in A} X_{\alpha}$.

• $\forall \alpha \in A, \emptyset \cap X_{\alpha} = \emptyset$, so $\emptyset \in \mathscr{T}$.
• $\forall \alpha \in A, (\sqcup_{\alpha \in A} X_{\alpha}) \cap X_{\alpha} = X_{\alpha}$, so $(\sqcup_{\alpha \in A} X_{\alpha}) \in \mathscr{T}$.
• Let $\{ U_{\beta} \} \subset \mathscr{T}$ where $B$ is the indexed set. Let $\alpha \in A$ be given. $\forall \beta, U_{\beta} \cap X_{\alpha} \in T_{\alpha}$. Then we have \((\bigcup_{\beta \in B} U_{\beta}) \cap X_{\alpha} = \bigcup_{\beta \in B} (U_{\beta} \cap X_{\alpha}) \in \mathscr{T}_{\alpha}\) since $\mathscr{T}_{\alpha}$ must be closed under arbitrary union. Since $\alpha$ was chosen arbitrarily, this implies that \(\bigcup_{\beta \in B} U_{\beta} \in \mathscr{T}\).
• Let $U_{\beta_1}, U_{\beta_2}, \cdots, U_{\beta_n} \in \mathscr{T}$. Let $\alpha \in A$ be given. $\forall i = 1, \cdots, n, U_{\beta_i} \cap X_{\alpha} \in T_{\alpha}$. Then we have \((\bigcap_{i=1}^n U_{\beta_i}) \cap X_{\alpha} = \bigcap_{i=1}^n (U_{\beta_i} \cap X_{\alpha}) \in \mathscr{T}_{\alpha}\) since $\mathscr{T}_{\alpha}$ must be closed under finite intersection. Since $\alpha$ was chosen arbitrarily, this implies that \(\bigcap_{i=1}^n U_{\beta_i} \in \mathscr{T}\).

Therefore, $\mathcal{T}$ is indeed a topology.