Exercise 19.3

Proof. Suppose that $x$ and $y$ are distinct points of $\prod <!--\; FUNCTION\; APPLICATION\; -->X_{\alpha }$. Then $x=(x_a)$ and $y=(y_{\alpha })$ where each $x_{\alpha },y_{\alpha }\in X_{\alpha }$, and there must be a $\beta$ where $x_{\beta }\ne y_{\beta }$ since $x\ne y$. Thus $x_{\beta }$ and $y_{\beta }$ are distinct points of $X_{\beta }$, so that there are neighborhoods $W_x$ and $W_y$ of $x_{\beta }$ and $y_{\beta }$, respectively, that are disjoint since $X_{\beta }$ is a Hausdorff space. So define the sets

so that clearly $x\in \prod <!--\; FUNCTION\; APPLICATION\; -->U_{\alpha }$ and $y\in \prod <!--\; FUNCTION\; APPLICATION\; -->V_{\alpha }$. Then since each $U_{\alpha }$ and $V_{\alpha }$ are open, we have that $\prod <!--\; FUNCTION\; APPLICATION\; -->U_{\alpha }$ and $\prod <!--\; FUNCTION\; APPLICATION\; -->V_{\alpha }$ are both basis elements of $\prod <!--\; FUNCTION\; APPLICATION\; -->X_{\alpha }$ and therefore open. Note that this is true for both the box and product topologies since, in the case of the latter, $U_{\alpha }$ and $V_{\alpha }$ are not all of $X_{\alpha }$ for only one $\alpha$, namely $\alpha =\beta$. Thus $\prod <!--\; FUNCTION\; APPLICATION\; -->U_{\alpha }$ is a neighborhood of $x$ and $\prod <!--\; FUNCTION\; APPLICATION\; -->V_{\alpha }$ is a neighborhood of $y$ in $\prod <!--\; FUNCTION\; APPLICATION\; -->X_{\alpha }$.

We also assert that $\prod <!--\; FUNCTION\; APPLICATION\; -->U_{\alpha }$ and $\prod <!--\; FUNCTION\; APPLICATION\; -->V_{\alpha }$ are disjoint, which of course completes the proof that $\prod <!--\; FUNCTION\; APPLICATION\; -->X_{\alpha }$ is Hausdorff. To see this, suppose to the contrary that there is a $z$ in both $\prod <!--\; FUNCTION\; APPLICATION\; -->U_{\alpha }$ and $\prod <!--\; FUNCTION\; APPLICATION\; -->V_{\alpha }$. Then $z=(z_{\alpha })$ and in particular we would have that $z_{\beta }\in U_{\beta }=W_x$ and $z_{\beta }\in V_{\beta }=W_y$. But then $z_{\beta }\in W_x\cap W_y$, which contradicts the fact that $W_x$ and $W_y$ are disjoint! So it must be that in fact $\prod <!--\; FUNCTION\; APPLICATION\; -->U_{\alpha }$ and $\prod <!--\; FUNCTION\; APPLICATION\; -->V_{\alpha }$ are disjoint. □