Proposition

Let $X$ be any space and $Y$ be a discrete space. Show that the Cartesian product $X \times Y$ is equal to the disjoint union $\coprod_{y \in Y} X$, and the product topology is the same as the disjoint union topology.

Solution

\[\begin{align*} \coprod_{y \in Y} X &= \bigcup_{y \in Y} X \times \{ y \} \\ &= \bigcup_{y \in Y} \{ x \times y \mid x \in X \} \\ &= \{ x \times y \mid x \in X, y \in Y \} \\ &= X \times Y. \end{align*}\]

Let $\mathscr{B}_X$ be a basis of $X$, $\mathscr{T}_X$ be the topology of $X$. Let $\mathscr{B}_Y = \{ \{ y \} \mid y \in Y \}$ be a basis of $Y$. Then $\mathscr{B}_X \times \mathscr{B}_Y$ is a basis for the product topology $\mathscr{T}_p$ of $X \times Y$.

Let $U_0 \times \{ y_0 \} \in \mathscr{B}_X \times \mathscr{B}_Y$ be given. Then $U_0 \times \{ y_0 \} = \coprod_{y \in Y} V_y$ where

\[\begin{align*} V_y &= \begin{cases} U_0 & (y = y_0) \\ \emptyset & (y \ne y_0). \end{cases} \end{align*}\]

For any $y \in Y$, the intersection of $\coprod_{y \in Y} V_y$ with $X$ (there is only one $X$, but the $X$ that corresponds to the $y$) is $V_y$. Since each $V_y$ is open in $X$, $\coprod_{y \in Y} V_y$ is open in the disjoint union topology. Therefore, this implies that the disjoint union topology is finer than the product topology.

Let $U \subset \coprod_{y \in Y} X$ be an open set. For each $y \in Y$, let $U_y$ denote the intersection of $U$ with $X$ for the index $y$. Each $U_y$ is open because $U$ is open. Then $U = \coprod_{y \in Y} U_y = \bigcup_{y \in Y} U_y \times \{ y \}$. Then each $U_y \times \{ y \}$ is a basis element in $\mathscr{B}_X \times \mathscr{B}_Y$. Therefore, $U$ is open in the product topology, so the product topology is finer than the disjoint union topology.

Hence, the product topology and the disjoint union topology are the same.