Let $\overline{A}$ define the closure of $A$. I'm asked to prove that
$$\overline{A}\cup\overline{B} = \overline{A \cup B}.$$
My attempt at this:
$\overline{A}\cup\overline{B}$ is the union of the smallest closed set containing $A$ and the smallest closed set containing $B$. Their union must obviously(?) be the smallest closed set containing $A$ and $B$, which exactly is the definition of $\overline{A\cup B}.$
Is this correct?
Best Answer
I don't think it is good to use the word "obviously" in that proof. Why is this the smallest set which contains $A$ and $B$? This is actually what you have to prove.
Anyway, it is easy. $\overline{A} \cup \overline{B}$ is a set which contains both $\overline{A}$ and $\overline{B}$. Since $A \subseteq \overline{A}$ and $B\subseteq \overline{B}$ we conclude $\overline{A}\cup\overline{B}$ contains $A \cup B$. And since $\overline{A}\cup\overline{B}$ is a closed set (as a finite union of closed sets) it must contain the closure of $A\cup B$. Hence $\overline{A\cup B}\subseteq \overline{A}\cup\overline{B}$.
As for the other direction, $\overline{A\cup B}$ is a closed set which contains $A\cup B$. So it contains $A$ and hence must contain $\overline{A}$. So $\overline{A}\subseteq\overline{A\cup B}$. For the same reasoning $\overline{B}\subseteq\overline{A\cup B}$. Hence $\overline{A}\cup\overline{B}\subseteq\overline{A\cup B}$.