ASpaceIsCompactIfAndOnlyIfTheSpaceHasTheFiniteIntersectionProperty 2003-04-12 14:28:20 2009-05-31 01:38:10 Theorem compact % 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 {\bf Theorem. } A topological space is compact if and only if any collection of its closed sets having the finite intersection property has non-empty intersection. The above theorem is essentially the definition of a compact space rewritten using de Morgan's laws. The usual definition of a compact space is based on open sets and unions. The above characterization, on the other hand, is written using closed sets and intersections. \emph{Proof.} Suppose $X$ is compact, i.e., any collection of open subsets that cover $X$ has a finite collection that also cover $X$. Further, suppose $\{F_i\}_{i\in I}$ is an arbitrary collection of closed subsets with the finite intersection property. We claim that $\cap_{i\in I} F_i$ is non-empty. Suppose otherwise, i.e., suppose $\cap_{i\in I} F_i=\emptyset$. Then, \begin{eqnarray*} X&=&\left(\bigcap_{i\in I} F_i\right)^c\\ &=&\bigcup_{i\in I} F_i^c. \end{eqnarray*} (Here, the complement of a set $A$ in $X$ is written as $A^c$.) Since each $F_i$ is closed, the collection $\{F_i^c\}_{i\in I}$ is an open cover for $X$. By compactness, there is a finite subset $J\subset I$ such that $X=\cup_{i\in J} F_i^c$. But then $X=(\cap_{i\in J} F_i)^c$, so $\cap_{i\in J} F_i=\emptyset$, which contradicts the finite intersection property of $\{F_i\}_{i\in I}$. The proof in the other direction is analogous. Suppose $X$ has the finite intersection property. To prove that $X$ is compact, let $\{F_i\}_{i\in I}$ be a collection of open sets in $X$ that cover $X$. We claim that this collection contains a finite subcollection of sets that also cover $X$. The proof is by contradiction. Suppose that $X\neq \cup_{i\in J} F_i$ holds for all finite $J\subset I$. Let us first show that the collection of closed subsets $\{F_i^c\}_{i\in I}$ has the finite intersection property. If $J$ is a finite subset of $I$, then \begin{eqnarray*} \bigcap_{i\in J} F^c_i &=& \Big(\bigcup_{i\in J} F_i\Big)^c \neq \emptyset, \end{eqnarray*} where the last assertion follows since $J$ was finite. Then, since $X$ has the finite intersection property, \begin{eqnarray*} \emptyset &\neq& \bigcap_{i\in I} F_i^c = \Big(\bigcup_{i\in I} F_i\Big)^c. \end{eqnarray*} This contradicts the assumption that $\{F_i\}_{i\in I}$ is a cover for $X$. $\Box$ \begin{thebibliography}{9} \bibitem{edwards} R.E. Edwards, \emph{Functional Analysis: Theory and Applications}, Dover Publications, 1995. \end{thebibliography}