Closure of the product of two sets vs product of their closures. This exercise is from Chapter 2, Section 17, number 9, pag. 101, from Munkres's Topology.
Let $A \subset X$ and $B \subset Y$. Show that in the space $X \times Y$, we have $\overline {A \times B} = \overline A \times \overline B$.
 A: Step 1. We begin by proving that $\overline A \times \overline B$ is a closed set. Because $\overline A$ is closed, then $X - \overline A$ is open in the space $X$; in the same way, as $\overline B$ is closed, then $Y - \overline B$ is open in the space $Y$. Therefore, the complement of $\overline A \times \overline B$ is the union of the sets $(X - \overline A) \times Y$ and $X \times (Y - \overline B)$, which are open in $X \times Y$, because they are the product of open sets. Since the union of open sets is an open set, the complement of $\overline A \times \overline B$ is an open set in the space $X \times Y$. It follows that $\overline A \times \overline B$ is closed.
Step 2. We have that $\overline A \times \overline B$ is a closed set that contains $A \times B$, because $A \subset \overline A$ and $B \subset \overline B$. But $\overline {A \times B}$ is the small closed set that contains $A \times B$. It follows that $\overline {A \times B} \subset \overline A \times \overline B$.
Step 3. Let $A^\prime$ be the set of all the ${\bf limit\, points}$ of $A$ in the space $X$. Let $B^\prime$ be the set of all the ${\bf limit\, points}$ of $B$ in the space $Y$. By Theorem 17.6, Pag. 97, from Munkres's book, we have $\overline A = A \bigcup A^\prime$ and $\overline B = B \bigcup B^\prime$. Then, it is easy to see that if $(x, y)$ is a point of $\overline A \times \overline B$, such that $x \in A^\prime$, then $(x, y)$ is a ${\bf limit\, point}$ of $A \times B$. On the other hand, if $(x, y)$ is a point of $\overline A \times \overline B$, such that $y \in B^\prime$, then $(x, y)$ is also a ${\bf limit\, point}$ of $A \times B$. We deduce that if $(x, y)$ is any point of $\overline A \times \overline B$, then $(x, y) \in A \times B$ or $(x, y)$ is a ${\bf limit\, point}$ of $A \times B$. This implies that $(x, y) \in \overline {A \times B}$, for which we deduce that $\overline A \times \overline B \subset \overline {A \times B}$.
By Step 2 and Step 3 we obtain $\overline {A \times B} = \overline A \times \overline B$.
