Some set notation clarification So for the product properties of sets, I can show that :
$(A \times B )^c=(X \times B^c)\cup (A^c\times Y)$, where $A\subset X$ and $B\subset Y,$ but I have a problem showing that $(A \times B )'=(\overline{A} \times B')\cup (A'\times \overline{B}),$ where the overline denotes the closure of the set, also note that $A'$ denotes the limit points of the set $A.$
Could anyone help me explain why the second equality holds? I have alot of topology question at the moment as my midterms are coming up, help would be much appreciated. 
 A: Suppose $A$ and $B$ are two topological spaces.
Then we need to proof $$(A \times B )'=(\overline{A} \times B')\cup (A'\times \overline{B})$$
Surfficiency: Suppose $x=(a,b) \in (A \times B )'$, then
$$x \in (A \times B )' \subset \overline{A \times B} = \overline{A} \times \overline{B}$$
that means $a \in \overline{A}$ and $b \in \overline{B}$, so we only need to prove $a \in A'$ or $b \in B'$, assume that $a \notin A'$ and $b \notin B'$, that means there exist a neighborhood $U_a$ of $a$ and a neighborhood $U_b$ of $b$, such that
$$\left(U_a - \{ {a} \}\right) \cap A=\emptyset\ \ \ \ and\ \ \ \  \left(U_b - \{ {b} \}\right) \cap B=\emptyset$$
hence the neighborhood $U_a \times U_b$ of  $x$ satisfies
$$ \left(U_a \times U_b - \{ {x} \}\right) \cap ( A \times B ) =\emptyset $$
therefore, $x \notin (A \times B )'$, it's a contradiction with the assume.
Necessary: Suppose $x \in (\overline{A} \times B')\cup (A'\times \overline{B})$, we need to proof that $ x \in (A \times B )'$. $\overline{A} \times B'$ are symmetrical to $A'\times \overline{B}$, so we can assume $x \in \overline{A} \times B'$ as well.
That means $a \in \overline{A}$ and $ b \in B'$ , for any neighborhood $U_x$ of $x$, we have 
$$U_x=\cup_{\mathcal{B}}(U_a \times U_b)$$ where $\mathcal{B}$ is a topological basis of $A \times B$. Choose an arbitrary $U_a \times U_b$ from $U_x=\cup_{\mathcal{B}}(U_a \times U_b)$, then either $a \in A$ or $a \in A'$, we have
$$b \in B' \Rightarrow (U_a \times U_b - \{x\})\cap(A \times B) \ne \emptyset$$
hence$$(U_x - \{x\})\cap(A \times B) \ne \emptyset\Rightarrow x \in (A \times B )'$$
The proof is complete. Furthur more, we have a similiar equality:
$$\partial(A \times B )=(\overline{A} \times \partial{B})\cup (\partial{A}\times \overline{B})$$
A: Have you tied sequences in product space $A\times B$ converging to a point (x,y) and considering the projections into $A$ and $B$?
$$(A \times B )'=(\overline{A} \times B')\cup (A'\times \overline{B})$$
