I'm having some trouble showing these two equal each other. I was doing some research and apparently this is not always true in a topological sense but I think that's out of the scope of this class.
$(A \cup B)' = A' \cup B'$
$A'$ is the set of all derived points of A.
Hints:
The easier containment is $A' \cup B' \subset (A \cup B)'$.
For the harder containment, suppose $x \in (A \cup B)'$. We need to show if $x \notin A'$ then $x \in B'$. Since $x \notin A'$ there exists an open set $U$ containing $x$ such that $A \cap U \subset \{x\}$. Let $V$ be any open set containing $x$ and consider that
1) $U \cap V$ is an open set containing $x$
2) $x \in (A \cup B)'$
and you should discover that $x \in B'$.
