Exercise 17.13

Proof. Suppose $\mathrm{\Delta }$ is closed in $X\times X$, i.e., the complement ${\mathrm{\Delta }}^c$ is open. This is equivalent to for all $(x,y)\in X\times X$ such that $x\ne y$, there exists a basis element $U\times V$ of $X\times X$ for $U,V$ open in $X$ such that $(x,y)\in U\times V$ but $(U\times V)\cap \mathrm{\Delta }=\varnothing$. But then, by definition of $\mathrm{\Delta }$, this is equivalent to saying for all $x,y\in X$ such that $x\ne y$, there exist open neighborhoods $U\ni x$ and $V\ni y$ such that $U\cap V=\varnothing$, and so $X$ is Hausdorff. □

Proof. $\text{(⇒)}$ Suppose that $X$ is Hausdorff and consider any point $x\times y\in X\times X$ where $x\times y\notin \mathrm{\Delta }$. Then it must be that $x\ne y$ so that there is a disjoint neighborhood $U$ of $x$ and $V$ of $y$ since $X$ is Hausdorff. Then $U\times V$ is a basis element of $X\times X$, by the definition of a product topology, and is therefore open. Now consider any point $w\times z\in U\times V$ so that $w\in U$ and $z\in V$. Then it has to be that $w\ne z$ since $U$ and $V$ are disjoint, which shows that $w\times z\notin \mathrm{\Delta }$. Since $w\times z$ was an arbitrary point of $U\times V$, this shows that $U\times V$ does not intersect $\mathrm{\Delta }$. Since also $U\times V$ is open and contains $x\times y$, this shows that $x\times y$ is not a limit point of $\mathrm{\Delta }$. Moreover, since $x\times y$ was an arbitrary element of $X\times X$ that is not in $\mathrm{\Delta }$, it follows that $\mathrm{\Delta }$ must contain all of its limit points and is therefore closed by Corollary 17.7.

$\text{(⇐)}$ Now suppose that $\mathrm{\Delta }$ is closed and suppose that $x$ and $y$ are distinct points in $X$. Then $x\times y\notin \mathrm{\Delta }$ so that $x\times y$ cannot be a limit point of $\mathrm{\Delta }$ (since it contains all its limit points by Corollary 17.7). Hence there is an open set $T$ in $X\times X$ that contains $x\times y$ and does not intersect $\mathrm{\Delta }$. It then follows that there is a basis element $U\times V$ of $X\times X$ containing $x\times y$ where $U\times V\subset T$. Then $U$ and $V$ are both open in $X$ by the definition of the product topology, and clearly $x\in U$ and $y\in V$. It also follows that $U\times V$ does not intersect $\mathrm{\Delta }$ since, if it did, then $T$ would as well.

Suppose that $U$ and $V$ are not disjoint so that there is a $z\in U$ where also $z\in V$. Then clearly $z\times z\in U\times V$ but we also have that $z\times z\in \mathrm{\Delta }$ so that $U\times V$ intersects $\mathrm{\Delta }$. As we know that this cannot be the case, it has to be that $U$ and $V$ are disjoint. This shows that $X$ is Hausdorff as desired since $U$ is a neighborhood of $x$, $V$ is a neighborhood of $y$, and $x$ and $y$ were arbitrary distinct points of $X$. □