TITLE: Closure of union of two sets
QUESTION [9 upvotes]: Show that
$$\overline{A\cup B} = \overline A\cup\overline B$$
$\overline A=A\cup A'$ where $A'$ are the limit points
My attempt: 
Since $\overline A$ is closed and $\overline B$, the union of two closed sets is closed.
Hence $\overline A\cup \overline B$ can be rewritten as $\overline{A\cup B}$
My Second attempt: 
$\overline{A\cup B}=(A\cup B)\cup (A\cup B)'=(A\cup B)\cup (A'\cup B')=A\cup A' \cup B \cup B' = \overline{A}\cup \overline{B}$

REPLY [2 votes]: Note that closures preserve set inclusion, i.e. if $A \subset B$ then $\overline A \subset \overline B$.
We want to show $\overline{A\cup B} = \overline{A} \cup \overline{B}$. $A \subset A\cup B$ and $B \subset A\cup B$ so $\overline{A} \subset \overline{A\cup B}$ and $\overline{B} \subset \overline{A\cup B}$. So, $\overline{A} \cup \overline{B} \subset \overline{A\cup B}$. On the other hand, $A \subset \overline{A}, B\subset \overline{B}$, so $A\cup B \subset \overline{A} \cup \overline{B}$. The latter is closed, so $\overline{A\cup B} \subset \overline{A} \cup \overline{B}$.
We conclude $\overline{A\cup B} = \overline{A} \cup \overline{B}$. $\ \  \blacksquare$