Exercise 17.11

Proof. Suppose that $X$ and $Y$ are Hausdorff spaces and consider two distinct points $(x_1,y_1)$ and $(x_2,y_2)$ in $X\times Y$. Since these points are distinct, it has to be that $x_1\ne x_2$ or $y_1\ne y_2$. In the first case, $x_1$ and $x_2$ are distinct points of $X$ so that there are disjoint neighborhoods $U_1$ and $U_2$ of $x_1$ and $x_2$, respectively. This of course follows from the fact that $X$ is a Hausdorff space. Then we have that $U_1\times Y$ and $U_2\times Y$ are both basis elements, and therefore open sets, in the product space $X\times Y$ since $Y$ itself is obviously an open set of $Y$. Clearly also $(x_1,y_1)\in U_1\times Y$ and $(x_2,y_2)\in U_2\times Y$ so that $U_1\times Y$ is a neighborhood of $(x_1,y_1)$ and $U_2\times Y$ is a neighborhood of $(x_2,y_2)$.

Then, for any $(x,y)\in U_1\times Y$ we have that $x\in U_1$ so that $x\notin U_2$ since they are disjoint. Then it has to be that $(x,y)\notin U_2\times Y$. This suffices to show that $U_1\times Y$ and $U_2\times Y$ are disjoint since $(x,y)$ was arbitrary. Thus $X\times Y$ is a Hausdorff space since the points $(x_1,y_1)$ and $(x_2,y_2)$ were arbitrary. An analogous argument in the case in which $y_1\ne y_2$ shows the same result. □