ProofOfPropertiesOfTheClosureOperator 2004-02-27 07:44:58 2004-02-27 07:44:58 Proof 0 % 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 \newtheorem{theorem}{Theorem} \newtheorem{defn}{Definition} \newtheorem{prop}{Proposition} \newtheorem{lemma}{Lemma} \newtheorem{cor}{Corollary} Recall that the closure of a set $A$ in a topological space $X$ is defined to be the intersection of all closed sets containing it. \begin{description} \item[$A\subset \overline{A}$]: By definition \[ \overline{A}=\bigcap_{C\supseteq A, \ C\text{ closed}} C, \] but since for every $C$ we have $A\subseteq C$, we immediately find \[ A\subseteq \bigcap_{C\supseteq A, \ C\text{ closed}} C. \] \item[$\overline{A}$ is closed]: Recall that the intersection of any number of closed sets is closed, so the closure is itself closed. \item[$\overline{\emptyset} = \emptyset$, $\overline{X} = X$, and $\overline{\overline{A}} = \overline{A}$]: If $C$ is any closed set, then \[ \overline{C} = \bigcap_{C'\supseteq C, \ C'\text{ closed}} C' = C \cap \bigcap_{C'\supsetneq C, \ C'\text{ closed}} C' = C. \] \item[$\overline{A\cup B} = \overline{A} \cup \overline{B}$]: First write down the definition: \begin{align*} \overline{A} \cup \overline{B} & = \bigcap_{C\supseteq A, \ C\text{ closed}} C \cup \bigcap_{D\supseteq B, \ D\text{ closed}} D, \\ \intertext{then apply DeMorgan's law to get} & = \bigcap_{C\supseteq A, D\supseteq B, \ C, D\text{ closed}}(C\cup D), \\ \intertext{but for every such pair $C$, $D$, we have that $E = C\cup D$ is a closed set containing $A\cup B$. Conversely, every closed set $E$ containing $A\cup B$ is obtained from such a pair --- just take $(E,E)$ to be the pair. Thus} & = \bigcap_{E\supseteq A\cup B, \ E\text{ closed}}(E) \\ & = \overline{A\cup B}. \end{align*} \item[$\overline{A\cap B} \subset \overline{A}\cap \overline{B}$]: \begin{align*} \overline{A} \cap \overline{B} & = \bigcap_{C\supseteq A, C\text{ closed}} C \cap \bigcap_{D\supseteq B, \ D\text{ closed}} D, \\ & = \bigcap_{C\supseteq A, D\supseteq B, \ C, D\text{ closed}}(C\cap D), \\ \intertext{but for every such pair $C$, $D$, we have that $E = C\cap D$ is a closed set containing $A\cap B$. However, some closed sets may not arise in this way, so we do not have equality. Thus} & \supseteq \bigcap_{E\supseteq A\cap B, \ E\text{ closed}}(E) \\ & = \overline{A\cap B}. \end{align*} so we have \[ \overline{A} \cap \overline{B} \supseteq \overline{A\cap B}. \] \item[$\overline{A} = A\cup A'$ where $A'$ is the set of all limit points of $A$]: Let $a$ be a limit point of $A$, and let $C$ be a closed set containing $A$. If $a$ is not in $C$, then $X\setminus C$ is an open set containing $a$ but not meeting $C$, which implies that $X\setminus C$ does not meet $A$, which contradicts the fact that $a$ was a limit point of $A$. Conversely, suppose that $a$ is not a limit point of $A$, and that $a$ is not in $A$. Then there is some open neighborhood $U$ of $a$ which does not meet $A$. But then $X\setminus U$ is a closed set containing $A$ but not containing $a$, so $a\notin\overline{A}$. \end{description}