# 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.

$\overline{\emptyset}=\emptyset$, $\overline{X}=X$, and $\overline{\overline{A}}=\overline{A}$

: If $C$ is any closed set, then

 $\overline{C}=\bigcap_{C^{\prime}\supseteq C,\ C^{\prime}\text{ closed}}C^{% \prime}=C\cap\bigcap_{C^{\prime}\supsetneq C,\ C^{\prime}\text{ closed}}C^{% \prime}=C.$
$\overline{A\cup B}=\overline{A}\cup\overline{B}$

: First write down the definition:

 $\displaystyle\overline{A}\cup\overline{B}$ $\displaystyle=\bigcap_{C\supseteq A,\ C\text{ closed}}C\cup\bigcap_{D\supseteq B% ,\ D\text{ closed}}D,$ 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 $\displaystyle=\bigcap_{E\supseteq A\cup B,\ E\text{ closed}}(E)$ $\displaystyle=\overline{A\cup B}.$
$\overline{A\cap B}\subset\overline{A}\cap\overline{B}$

:

 $\displaystyle\overline{A}\cap\overline{B}$ $\displaystyle=\bigcap_{C\supseteq A,C\text{ closed}}C\cap\bigcap_{D\supseteq B% ,\ D\text{ closed}}D,$ $\displaystyle=\bigcap_{C\supseteq A,D\supseteq B,\ C,D\text{ closed}}(C\cap D),$ 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 $\displaystyle\supseteq\bigcap_{E\supseteq A\cap B,\ E\text{ closed}}(E)$ $\displaystyle=\overline{A\cap B}.$

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 ProofOfPropertiesOfTheClosureOperator 2013-03-22 14:12:19 2013-03-22 14:12:19 archibal (4430) archibal (4430) 4 archibal (4430) Proof msc 54A99