I need to prove that $$\text{Fr}(A\cup B) = \text{Fr}(A) \cup \text{Fr}(B)$$
if $\overline{A}\cap\overline{B}=\emptyset$.
I have been able to prove the left to right inclusion ($\subseteq$).
But what about the the other inclusion ($\supseteq$)?
my attempt :
$$\text{Fr}(A\cup B) \cup \text{Fr}(A\cap B)= \text{Fr}(A) \cup \text{Fr}(B).$$
So if $ \overline{A}\cap\overline{B}=\emptyset$ implies that $A\cap B= \emptyset$ and since $\text{Fr}(\emptyset)=\emptyset$ we conclude.
Is this correct?
Best Answer
You proof is correct if you assume true (without proof)) your formula. I refer to a similar approach in this nice answer. Note that with your last approach you are proving equality immediately: you would not even need the first inclusion.
Here is another (slightly more general approach): For the last implication, it is not even necessary to ask $\overline{A} \cap \overline{B} = \emptyset$. If $\overline{A} \cap B = \emptyset = A \cap \overline{B}$, the result holds.
You already proved that $\partial ( A \cup B ) \subseteq \partial A \cup \partial B$.
To prove the converse (by contradiction) assume that $x \notin \partial ( A \cup B )$. Then you can distinguish two possibilities: