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$$ \overline{C} = \bigcap_{C'\supseteq C, \ C'\text{ closed}} C' = C \cap \bigcap_{C'\supsetneq C, \ C'\text{ closed}} C' = C.$$

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

: First write down the definition:

then apply DeMorgan's law to get
but for every such pair , , we have that is a closed set containing . Conversely, every closed set containing is obtained from such a pair -- just take to be the pair. Thus

$\overline{A\cap B} \subset \overline{A}\cap \overline{B}$

:

but for every such pair , , we have that is a closed set containing . However, some closed sets may not arise in this way, so we do not have equality. Thus

so we have$$ \overline{A} \cap \overline{B} \supseteq \overline{A\cap B}.$$

$\overline{A} = A\cup A'$ 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 $a\notin\overline{A}$ .