ProductTopologyPreservesTheHausdorffProperty 2003-05-31 05:13:21 2004-03-12 09:59:49 Theorem product topology % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \newcommand{\sR}[0]{\mathbb{R}} \newcommand{\sC}[0]{\mathbb{C}} \newcommand{\sN}[0]{\mathbb{N}} \newcommand{\sZ}[0]{\mathbb{Z}} {\bf 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. \emph{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 \PMlinkname{here}{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 \PMlinkname{preimage commutes with set operations}{InverseImageCommutesWithSetOperations}, 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. $\Box$