Separating Closed Sets in a Well-Ordered Set in the Order Topology

Question

Theorem 32.4 from Munkres' Topology book:

Every well-ordered set $X$ is normal in the order topology

This is something a continuation of this post, where, evidently, I forgot what the definition of a well-ordered space. In this post, I am trying to prove the following claim Munkres uses in his proof of theorem 32.4:

Let $B$ be closed in $X$, and $a \in X-B$. Then there exists a basis element about $a$ which is disjoint from $B$ and contains $(x,a]$ for some $x \in X$.

I have tried to prove this, but I don't know how to. I could use some help.

Answer 1

$B$ is closed, so $X\setminus B$ is open and contains $a$. By the fact (linked to by you) every set of the form $(x,a]$ where $x < a$ is open and these clearly form a local base at $a$ (this follows from the definition of the order topology). So one of these is contained in $X\setminus B$, so there is some $(x,a]$ that is disjoint from $B$, as claimed.

Only when $a = \min(X)$ this will fail (this is a special case); in that case for some $x$, $[a,x)$ is open and disjoint from $B$. But Munkres excludes the minimal element from his proof at first for exactly this reason. He can later deal with this isolated point later (as $\{\min(X)\} = [\min(X), \min(X)^+)$ is open in the order topology on a well ordered set).

Answer 2

As $B$ is closed, $X-B$ is open, so you immediately get a basis element that contains $a$ and is disjoint from $B$. That this basis element includes $(x,a]$ for some $x\in X$ is undoubtedly not what Munkres meant, since it's trivially true with $x=a$. A more reasonable version would be that it contains $(x,a]$ for some $x<a$; that's true unless $a$ happens to be the first element of $X$.