proof of properties of the closure operator
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.
 $A\subset \overline{A}$

: 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
$$A\subseteq \bigcap _{C\supseteq A,C\text{closed}}C.$$  $\overline{A}$ is closed

: Recall that the intersection of any number of closed sets is closed, so the closure is itself closed.
 $\overline{\mathrm{\varnothing}}=\mathrm{\varnothing}$, $\overline{X}=X$, and $\overline{\overline{A}}=\overline{A}$

: If $C$ is any closed set, then
$$\overline{C}=\bigcap _{{C}^{\prime}\supseteq C,{C}^{\prime}\text{closed}}{C}^{\prime}=C\cap \bigcap _{{C}^{\prime}\u228bC,{C}^{\prime}\text{closed}}{C}^{\prime}=C.$$  $\overline{A\cup B}=\overline{A}\cup \overline{B}$

: First write down the definition:
$\overline{A}\cup \overline{B}$ $={\displaystyle \bigcap _{C\supseteq A,C\text{closed}}}C\cup {\displaystyle \bigcap _{D\supseteq B,D\text{closed}}}D,$ then apply DeMorgan’s law to get $={\displaystyle \bigcap _{C\supseteq A,D\supseteq B,C,D\text{closed}}}(C\cup D),$ 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 $={\displaystyle \bigcap _{E\supseteq A\cup B,E\text{closed}}}(E)$ $=\overline{A\cup B}.$  $\overline{A\cap B}\subset \overline{A}\cap \overline{B}$

:
$\overline{A}\cap \overline{B}$ $={\displaystyle \bigcap _{C\supseteq A,C\text{closed}}}C\cap {\displaystyle \bigcap _{D\supseteq B,D\text{closed}}}D,$ $={\displaystyle \bigcap _{C\supseteq A,D\supseteq B,C,D\text{closed}}}(C\cap D),$ 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 $\supseteq {\displaystyle \bigcap _{E\supseteq A\cap B,E\text{closed}}}(E)$ $=\overline{A\cap B}.$ so we have
$$\overline{A}\cap \overline{B}\supseteq \overline{A\cap B}.$$  $\overline{A}=A\cup {A}^{\prime}$ where ${A}^{\prime}$ 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 $a\notin \overline{A}$.
Title  proof of properties of the closure operator 

Canonical name  ProofOfPropertiesOfTheClosureOperator 
Date of creation  20130322 14:12:19 
Last modified on  20130322 14:12:19 
Owner  archibal (4430) 
Last modified by  archibal (4430) 
Numerical id  4 
Author  archibal (4430) 
Entry type  Proof 
Classification  msc 54A99 