Is it true that $\overline{A}\cup\overline{B}$ = $\overline{A\cup B}$? $\overline{A}$ denotes the closure of the set $A$.
I can show that $\overline{A}\cup\overline{B}\subset \overline{A\cup B}$, but I don't know how to go about the reverse direction?
 A: $\overline{A} \cup \overline{B}$ is a closed set containing $A \cup B$. Hence $\overline{A \cup B} \subseteq \overline{A} \cup \overline{B}$.
A: Consider a point $x$ in $\overline{A\cup B} $. Then for any nbhd $U$ of $x$,  $U\cap (A\cup B)\neq\emptyset $. So for any nbhd $U$ of $x, U\cap A\neq \emptyset$ or $U\cap B\neq \emptyset$.  Hence $x\in\overline A$ or $x\in\overline B$ and we are done. 
A: Here's an argument for first-countable spaces (e.g. $\mathbb{R}^n$):
It suffices to show that a limit point of $A \cup B$ is either a limit point of $A$ or a limit point of $B$.  If $x$ is a limit point of $A \cup B$, then this means we can find a sequence $\{x_k\}_{k=1}^\infty$ that converges to $x$ where $x_k \in A$ or $x_k \in B$ for each $k$.  This sequence admits a subsequence $\{x_{k_m}\}_{m=1}^\infty$ of points contained entirely in $A$ or entirely in $B$.  Since subsequences of convergent sequences also converge to the same point, we therefore have a sequence of points in either $A$ or $B$ that converges to $x$.  Hence, $x \in \overline{A}$ or $x \in \overline{B} \implies x \in \overline{A} \cup \overline{B} \implies \overline{A} \cup \overline{B} \subset \overline{A \cup B}$.
A: hint
$$A\subset A\cup B \implies $$
$$\overline{A} \subset \overline{A\cup B} $$
by the same, $$\overline {B}\subset \overline {A\cup B} $$
done.
A: Let $p\in \overline{A\cup B}$.  Then $p\in A\cup B$ or $p$ is an accumulation point of $A\cup B$.  We proceed by cases.
Assume first that $p\in A\cup B$, then $p\in A$ or $p\in B$.  If $p\in A$, then $p\in\overline{A}$ so $p\in\overline{A}\cup\overline{B}$.  If $p\in B$, then $p\in\overline{B}$, so $p\in\overline{A}\cup\overline{B}$.  Therefore, $p\in\overline{A}\cup\overline{B}$.
Now, assume $p$ is an accumulation point of $A\cup B$, but $p\not\in A\cup B$.  Assume, for contradiction, that $p$ is not an accumulation point of $A$ and not an accumulation point of $B$.  Then, there are open sets $S_1$ and $S_2$ containing $p$ such that $S_1\cap A=\emptyset$ and $S_2\cap B=\emptyset$ (these are empty sets because the intersection can only be $p$ and $p\not\in A\cup B$).
Let $S=S_1\cap S_2$.  This is an open set containing $p$ and $S\cap A=\emptyset=S\cap B$.  Therefore, $S\cap (A\cup B)=\{p\}$, so $p$ is not an accumulation point of $A\cup B$, a contradiction.
Now, we know that $(\overline{A\bigcup B})\subset \overline{A}\bigcup \overline{B}$. Let $p\in \overline{A}\bigcup \overline{B}$, then $p$ is an accumulation point of $A$ or $p$ is an accumulation point of $B$, or both. Then, $p$ is certainly in $(\overline{A\bigcup B})$ which implies $\overline{A}\bigcup \overline{B} \subset (\overline{A\bigcup B})$.
