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

$$\overline{C}=\bigcap _{C^{\prime }\supseteq C,C^{\prime }\text{ closed}}{C}^{\prime }=C\cap \bigcap _{C^{\prime }⊋C,C^{\prime }\text{ closed}}{C}^{\prime }=C.$$

$\overline{A\cup B}=\overline{A}\cup \overline{B}$

: 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 $a\notin \overline{A}$.