Exercise 17.10

Proof. Suppose that $X$ is an ordered set with the order topology. Consider a pair of distinct points $x_1$ and $x_2$ in $X$. Since $X$ is an order, $x_1$ and $x_2$ must be comparable since they are distinct, so we can assume that $x_1<x_2$ without loss of generality.

Case: $x_2$ is the immediate successor of $x_1$. Then, if $X$ has a smallest element $a$ then clearly the set $U_1=[a,x_2)$ is a neighborhood (because it is a basis element) of $x_1$. If $X$ has no smallest element then there is an $a<x_1$ so that $U_1=(a,x_2)$ is a neighborhood of $x_1$. Similarly $U_2=(x_1,b]$ or $U_2=(x_1,b)$ is a neighborhood of $x_2$, where $b$ is either the largest element of $X$ or $x_2<b$, respectively. Either way, for any $y\in U_1$ we have that $y<x_2$ so that $y\le x_1$ since $x_2$ is the immediate successor of $x_1$. Hence it is not true that $y>x_1$ so that $y\notin U_2$. This shows that $U_1$ and $U_2$ are disjoint.

Case: $x_2$ is not the immediate successor of $x_1$. Then there is an $x\in X$ where $x_1<x<x_2$. So let $U_1=[a,x)$ (or $U_1=(a,x)$) for the smallest element $a$ of $X$ (or some $a<x_1$). Similarly let $U_2=(x,b]$ (or $U_2=(x,b)$) for the largest element $b$ of $X$ (or some $x_2<b$). Either way $U_1$ and $U_2$ are neighborhoods of $x_1$ and $x_2$, respectively. If $y\in U_1$ then $y<x$ so that clearly it is not true that $y>x$ so that $x\notin U_2$. Hence again $U_1$ and $U_2$ are disjoint.

Thus in either case we have shown that $X$ is a Hausdorff space as desired since $x_1$ and $x_2$ were an arbitrary pair. □