ConditionsForACollectionOfSubsetsToBeABasisForSomeTopology 2004-05-10 19:17:08 2004-05-10 19:17:08 Proof basis (topology) 0 teaching proofs characterization of a basis what a basis looks like % 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 \def\co{\colon\thinspace} \theoremstyle{definition} \newtheorem*{thm}{Theorem} Not just any collection of subsets of $X$ can be a basis for a topology on $X$. For instance, if we took $\mathcal{C}$ to be all open intervals of length $1$ in $\mathbb{R}$, $\mathcal{C}$ isn't the basis for any topology on $\mathbb{R}$: $(0,1)$ and $(.5, 1.5)$ are unions of elements of $\mathcal{C}$, but their intersection $(.5,1)$ is not. The collection formed by arbitrary unions of members of $\mathcal{C}$ isn't closed under finite intersections and isn't a topology. We'd like to know which collections $\mathcal{B}$ of subsets of $X$ could be the basis for some topology on $X$. Here's the result: \begin{thm} A collection $\mathcal{B}$ of subsets of $X$ is a basis for some topology on $X$ if and only if: \begin{enumerate} \item Every $x\in X$ is contained in some $B_x\in \mathcal{B}$, and \item If $B_1$ and $B_2$ are two elements of $\mathcal{B}$ containing $x\in X$, then there's a third element $B_3$ of $\mathcal{B}$ such that $x\in B_3\subset B_1\cap B_2$. \end{enumerate} \end{thm} \begin{proof} First, we'll show that if $\mathcal{B}$ is the basis for some topology $\mathcal{T}$ on $X$, then it satisfies the two conditions listed. $\mathcal{T}$ is a topology on $X$, so $X\in \mathcal{T}$. Since $\mathcal{B}$ is a basis for $\mathcal{T}$, that means $X$ can be written as a union of members of $\mathcal{B}$: since every $x\in X$ is in this union, every $x\in X$ is contained in some member of $\mathcal{B}$. That takes care of the first condition. For the second condition: if $B_1$ and $B_2$ are elements of $\mathcal{B}$, they're also in $\mathcal{T}$. $\mathcal{T}$ is closed under intersection, so $B_1\cap B_2$ is open in $\mathcal{T}$. Then $B_1\cap B_2$ can be written as a union of members of $\mathcal{B}$, and any $x\in B_1\cap B_2$ is contained by some basis element in this union. Second, we'll show that if a collection $\mathcal{B}$ of subsets of $X$ satisfies the two conditions, then the collection $\mathcal{T}$ of unions of members of $\mathcal{B}$ is a topology on $X$. \begin{itemize} \item $\emptyset \in \mathcal{T}$: $\emptyset$ is the null union of zero elements of $\mathcal{B}$. \item $X\in \mathcal{T}$: by the first condition, every $X$ is contained in some member of $\mathcal{B}$. The union of all the members of $\mathcal{B}$ is then all of $X$. \item $\mathcal{T}$ is closed under arbitrary unions: Say we have a union of sets $T_{\alpha}\in \mathcal{T}$... \begin{align*} \bigcup_{\alpha \in I} T_{\alpha} &= \bigcup_{\alpha \in I} \bigcup_{\beta \in J_{\alpha}} B_{\beta} \\ \intertext{(since each $T_{\alpha}$ is a union of sets in $\mathcal{B}$)} &= \bigcup_{\beta \in \bigcup_{\alpha \in I} J_{\alpha}} B_{\beta} \end{align*} Since that's a union of elements of $\mathcal{B}$, it's also a member of $\mathcal{T}$. \item $\mathcal{T}$ is closed under finite intersections: since a collection of sets is closed under finite intersections if and only if it is closed under pairwise intersections, we need only check that the intersection of two members $T_1, T_2$ of $\mathcal{T}$ is in $\mathcal{T}$. Any $x\in T_1\cap T_2$ is contained in some $B_x^1\subset T_1$ and $B_x^2\subset T_2$. By the second condition, $x\in B_x^1\cap B_x^2$ gets us a $B_x^3$ with $x\in B_x^3 \subset B_x^1\cap B_x^2 \subset T_1\cap T_2$. Then \[ T_1\cap T_2 = \bigcup_{x\in T_1\cap T_2} B_x^3 \] which is in $\mathcal{T}$. \end{itemize} \end{proof}