Is my proof that $\overline{A}\cup\overline{B} = \overline{A\cup B}$ correct? 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? 
 A: 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 $\overline{A}\cup\overline{B}\subseteq\overline{A\cup B}$. 
A: No, since there's no proof. You just claim that the result you want to prove is obvious.
A: 1)$A \subset \overline {A}$, $B \subset \overline{B}$;
$A\cup B \subset \overline{A} \cup \overline{B}$ , the latter  as a union of $2$ closed sets is closed.
Hence 
$\overline {A\cup B} \subset \overline{A} \cup \overline {B}$, since $\overline {A\cup B}$ is the smallest closed set with $A \cup B \subset \overline{A\cup B}$.
2) Still need to show:
$\overline {A} \cup \overline {B} \subset \overline {A \cup B}.$
A: The closure operator $\mathrm{cl}$ satisfies some nice properties, namely,


*

*$A\subseteq B\implies \mathrm{cl}(A)\subseteq \mathrm{cl}(B)$

*$A\subseteq \mathrm{cl}(A)$

*$\mathrm{cl}(\mathrm{cl}(A))=\mathrm{cl}(A)$
To prove the equality $\mathrm{cl}(A)\cup\mathrm{cl}(B)=\mathrm{cl}(A\cup B)$, you must show both inclusions. One inclusion is easy because of the above properties. Indeed, we have $\mathrm{cl}(A)\subseteq \mathrm{cl}(A\cup B)$ and $\mathrm{cl}(B)\subseteq \mathrm{cl}(A\cup B)$ by the first bullet point. Therefore $$\mathrm{cl}(A)\cup\mathrm{cl}(B)\subseteq \mathrm{cl}(A\cup B).$$
It remains to show the other inclusion. This will not follow from the above properties, and is a special property when dealing with topological closures. I recommend that you use the definition of $\mathrm{cl}$ to proceed.
