box topology BoxTopology 2002-06-12 02:52:42 2006-09-10 03:50:10 Definition box product \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} \def\T{{\mathcal T}} \def\B{{\mathcal B}} \def\S{{\mathcal S}} \def\bigtimes{\mathop{\mbox{\Huge $\times$}}} \PMlinkescapeword{index} Let $\{ (X_\alpha,\T_\alpha) \}_{\alpha\in A}$ be a family of topological spaces. Let $Y$ denote the generalized Cartesian product of the sets $X_\alpha$, that is \[ Y = \prod_{\alpha\in A} X_\alpha. \] Let $\B$ denote the set of all products of open sets of the corresponding spaces, that is \[ \B = \left\{ \prod_{\alpha\in A} U_\alpha \,\Biggm|\, U_\alpha\in\T_\alpha \text{ for all } \alpha\in A \right\}. \] Now we can construct the \emph{box product} $(Y,\S)$, where $\S$, referred to as the {\em box topology}, is the topology \PMlinkescapetext{generated by} the base $\B$. When $A$ is a \PMlinkname{finite}{Finite} set, the box topology coincides with the product topology. \section*{Example} As an example, the box product of two topological spaces $(X_0,\T_0)$ and $(X_1,\T_1)$ is $(X_0\times X_1,\S)$, where the box topology $\S$ (which is the same as the product topology) consists of all sets of the form $\bigcup_{i\in I}(U_i\times V_i)$, where $I$ is some index set and for each $i\in I$ we have $U_i\in\T_0$ and $V_i\in\T_1$.