# product topology preserves the Hausdorff property

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 (http://planetmath.org/GeneralizedCartesianProduct). 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 (http://planetmath.org/InverseImageCommutesWithSetOperations), we have that

 $\displaystyle\pi_{\beta}^{-1}(U)\cap\pi_{\beta}^{-1}(V)$ $\displaystyle=$ $\displaystyle\pi_{\beta}^{-1}\big{(}U\cap V\big{)}$ $\displaystyle=$ $\displaystyle\emptyset.$

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. $\Box$

Title product topology preserves the Hausdorff property ProductTopologyPreservesTheHausdorffProperty 2013-03-22 13:39:40 2013-03-22 13:39:40 archibal (4430) archibal (4430) 7 archibal (4430) Theorem msc 54B10 msc 54D10