Exercise 17.9

Proof. $\text{(⊂)}$ Consider $(x,y)\in \overline{A\times B}$. Also suppose that $U$ and $V$ are any open sets in $X$ and $Y$, respectively, that contain $x$ and $y$, respectively. Then $U\times V$ is a basis element of the product topology on $X\times Y$, by definition, that contains $(x,y)$. It then follows from Theorem 17.5 part (b) that $U\times V$ intersects $A\times B$ and hence there is a point $(w,z)\in U\times V$ where also $(w,z)\in A\times B$. Then $w\in U$ and $w\in A$ so that $U$ intersects $A$, and hence $x\in \overline{A}$ by Theorem 17.5 part (a) since $U$ was an arbitrary neighborhood of $x$. An analogous argument shows that $y\in \overline{B}$. Therefore $(x,y)\in \overline{A}\times \overline{B}$ so that $\overline{A\times B}\subset \overline{A}\times \overline{B}$ since $x$ was arbitrary.

$\text{(⊃)}$ Now suppose that $(x,y)$ is any point in $\overline{A}\times \overline{B}$ so that $x\in \overline{A}$ and $y\in \overline{B}$. Suppose also that $U\times V$ is any basis element of $X\times Y$ that contains $(x,y)$ so that by definition $U$ and $V$ are open in $X$ and $Y$, respectively. Since $x\in \overline{A}$ and $U$ is an open set where $x\in U$, it follows from Theorem 17.5 part (a) that $U$ intersects $A$. Thus there is $w\in U$ where $w\in A$ as well. An analogous argument shows that $V$ intersects $B$ so that there is a $z\in V$ where also $z\in B$. We, therefore, have that $(w,z)\in U\times V$ and $(w,z)\in A\times B$ so that $U\times V$ intersects $A\times B$. Since $U\times V$ was an arbitrary basis element containing $(x,y)$, it follows from Theorem 17.5 part (b) that $(x,y)\in \overline{A\times B}$. This shows that $\overline{A}\times \overline{B}\subset \overline{A\times B}$ since the point $(x,y)$ was arbitrary. □