Show that if $A$ and $B$ are subsets of a set $S$, then $\overline{A \cap B}=\overline{A}\cup \overline{B}$.
I tried to prove that $A \cap B=A \cup B$ because I didn't realize that the overline meant to prove it for the closure of the sets.
So, now I am confused about how to prove for closure. I cannot find it in my textbook, and by some "similar" proofs online led me to conclude that $\overline{A \cap B}=\overline{A \cup B}$ but I somehow don't know if this is true, or how to prove it exactly. So, now I am not sure if I understand this principle at all.
Best Answer
Like I said in my comment, I'm pretty sure that $\overline A$ is referring to the complement of $A$ in $S$. The way to prove this problem is to just blindly "chase elements":
Let $x\in\overline{A\cap B}$. Then $x\in S$ but $x\notin A\cap B$. Therefore $x\notin A$ or $x\notin B$. This precisely means $x\in\overline A\cup\overline B$, so $\overline{A\cap B}\subseteq\overline A\cup\overline B$.
I would encourage you do the other direction on your own. Just follow the same procedure I did above, and follow the definitions to show $\overline A\cup\overline B\subseteq\overline{A\cap B}$.