product topology preserves the Hausdorff property

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 $U\cap V=\mathrm{\varnothing }$. 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

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

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

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

Title	product topology preserves the Hausdorff property
Canonical name	ProductTopologyPreservesTheHausdorffProperty
Date of creation	2013-03-22 13:39:40
Last modified on	2013-03-22 13:39:40
Owner	archibal (4430)
Last modified by	archibal (4430)
Numerical id	7
Author	archibal (4430)
Entry type	Theorem
Classification	msc 54B10
Classification	msc 54D10