product topology preserves the Hausdorff property

Theorem Suppose ${\{X_{\alpha }\}}_{\alpha \in A}$ is a collection ofHausdorff spaces. Then thegeneralized Cartesian product${\prod }_{\alpha \in A}{X}_{\alpha }$equipped with the product topology is a Hausdorff space.

Proof. Let $Y={\prod }_{\alpha \in A}{X}_{\alpha }$, andlet $x,y$ be distinct points in $Y$. Then there is an index $\beta \in A$such that $x(\beta )$ and $y(\beta )$ are distinct points inthe 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 $U\cap V=\mathrm{\varnothing }$.Let ${\pi }_{\beta }$ be the projection operator $Y\to X_{\beta }$ definedhere (http://planetmath.org/GeneralizedCartesianProduct). By the definition ofthe product topology, ${\pi }_{\beta }$ is continuous, so${\pi }_{\beta }^{-1}(U)$ and ${\pi }_{\beta }^{-1}(V)$ are open sets in $Y$. Also,since thepreimage commutes with set operations (http://planetmath.org/InverseImageCommutesWithSetOperations),we have that

${\pi }_{\beta }^{-1}(U)\cap {\pi }_{\beta }^{-1}(V)$$=$${\pi }_{\beta }^{-1}\left(U\cap V\right)$

$=$$\mathrm{\varnothing }.$

Finally, since $x(\beta )\in U$, i.e., ${\pi }_{\beta }(x)\in U$,it followsthat $x\in {\pi }_{\beta }^{-1}(U)$. 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.$\mathrm{\square }$