Am reading Munkres' Topology book Theorem 32.4: Every well-ordered set $X$ is normal in the order topology.
Munkres first shows that if $A$ and $B$ are disjoints closed sets in $X$, and neither $A$ nor $B$ contains the smallest element $a_0$ of $X$, then there exist disjoints open sets containing $A$ and $B$ respectively.
He then writes:
Finally, assume $A$ and $B$ are disjoints closed sets in $X$, and $A$ contains the smallest element $a_0$ of $X$. The set $\{a_0\}$ is both open and closed in $X$ (see here). By the result of the preceding paragraph, there exist open sets $U$ and $V$ containing the closed sets $A \setminus\{a_0\}$ and $B$, respectively. Then $U\cup \{a_0\}$ and $V$ are disjoints open sets containing $A$ and $B$, respectively.
Question: Shoudn't we consider $U\cup \{a_0\}$ and $V\setminus \{a_0\}$ instead?
Best Answer
Indeed it would be more precise to consider $U \cup \{a_0\}$ and $V\setminus \{a_0\}$. These are still disjoint and both open as well.