Exercise 31.3

Proof. Let $X$ be an ordered set with the order topology. $X$ is Hausdorff and therefore $T_1$ by Theorem $17.11$. It therefore suffices to show the condition in Lemma $31.1(a)$. So let $x\in X$ and let $U$ be an open neighborhood of $x$; we will construct a basis element $V$ of the order topology such that $x\in V$ and $\overline{V}\subseteq U$.

Suppose $x$ is neither the smallest nor largest element of $X$. Then, $x\in (a,b)\subseteq U$ for some basis element $(a,b)$ of the order topology. If $(a,x),(x,b)$ are nonempty, then let $V=(u,v)$ where $u\in (a,x)$ and $v\in (x,b)$. If $(a,x)=\varnothing$, let $u=a$, so that $V=(u,v)=[x,v)$; if $(x,b)=\varnothing$, let $v=b$, so that $V=(u,v)=(u,x]$. Then, $x\in (u,v)$ and $\overline{(u,v)}\subseteq (a,b)\subseteq U$.

Now suppose $x$ is the smallest (resp. largest) element of $X$. Then, $x\in [x,b)\subseteq U$ (resp. $x\in (a,x]\subseteq U$) for some basis element $[x,b)$ (resp. $(a,x]$) of the order topology. Now if $(x,b)$ (resp. $(a,x)$) is nonempty, then let $V=[x,v)$ (resp. $(u,x]$) where $v\in (x,b)$ (resp. $u\in (a,x)$). On the other hand, if $(x,b)$ (resp. $(a,x)$) is empty, then let $v=b$ (resp. $u=a$), so that $V=\{x\}$. We then have $x\in V$ and $\overline{V}\subseteq U$.

If $x$ is the smallest and largest element of $X$, then $X=\{x\}$ is trivially regular. □