Theorem Suppose $\{X_\alpha\}_{\alpha\in A}$ is a collection of Hausdorff spaces. Then the generalized Cartesian product $ \prod_{\alpha\in A} X_\alpha $ equipped with the product topology is a Hausdorff space.

Proof. Let $Y=\prod_{\alpha\in A} X_\alpha$ , and let $x,y$ be distinct points in $Y$ . Then there is an index $\beta \in A$ such that $x(\beta)$ and $y(\beta)$ are distinct points in the Hausdorff space $X_\beta$ . It follows that there are open sets $U$ and $V$ in $X_\beta$ such that $x(\beta)\in U$ , $y(\beta) \in V$ , and $U\cap V = \emptyset$ . Let $\pi_\beta$ be the projection operator $Y\to X_\beta$ defined here. By the definition of the product topology, $\pi_\beta$ is continuous, so $\pi_\beta^{-1}(U)$ and $\pi_\beta^{-1}(V)$ are open sets in $Y$ . Also, since the preimage commutes with set operations, we have that \begin{eqnarray*} \pi_\beta^{-1}(U) \cap \pi_\beta^{-1}(V) &=& \pi_\beta^{-1} \big(U \cap V\big) \\ &=& \emptyset. \end{eqnarray*}Finally, since $x(\beta)\in U$ , i.e., $\pi_\beta(x)\in U$ , it follows that $x\in \pi_\beta^{-1}(U)$ . Similarly, $y\in \pi_\beta^{-1}(V)$ . We have shown that $U$ and $V$ are open disjoint neighborhoods of $x$ respectively $y$ . In other words, $Y$ is a Hausdorff space.