Adherence and closed sets

Question

Please How to prove that if $A\cap \overline{B}=\overline{A}\cap B=\emptyset$ and if $A\cup B$ is closed then $A$ and $B$ are closed .

We deduce from $A\cap \overline{B}=\overline{A}\cap B=\emptyset$ than

$$\overline{A}\subset [X\setminus B],~ B\subset [X\setminus\overline{A}]\\~\text{and}\\ A\subset [X\setminus\overline{B}],~ \overline{B}\subset [X\setminus A] $$

but i don't know how to use this with $\overline{A\cup B}=\overline{A}\cup\overline{B}=A\cup B$ to find that $\overline{A}=A$ and $\overline{B}=B$

Answer 1

You know that $A\cup B $ is closed and $\bar{B}$ is closed. Thus $$(A\cup B)\cap \bar{B} = (A\cap \bar{B})\cup(B\cap \bar{B})=B$$ is closed. Hence $B= \bar{B}$. $A$ follows similarly.

Answer 2

Im gonna prove A is closed, (B is closed its the same reasoning)

$A∪B$ is closed: so $(A∪B)^c$ is an open set.

then $A^c∩B^c$=$(A∪B)^c$ is an open set.

lets say $x∈A^c$, then $x∈A^c∩B^c$ or $x∈A^c∩B$

if its the first case: Then golden, since $A^c∩B^c$ is an open set you have a neighbourhood of $x$ included in $A^c$

if is the second case, then $x∈B$ but, since $\overline{A}\cap B=\emptyset$ then $x$ has to have a neigborhood included in $A^c$.

Then for every element $x$ in $A^c$ you have a neighborhood $V$ of $x$ included in $A^c$, consequently $A^c$ is an open set. Then, by definition, $A$ has to be a closed set.

Hope it helps.