TITLE: No Immediate Successor? No Largest Element?
QUESTION [6 upvotes]: I am reading through the proof of the following theorem in Munkres' topology:

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

Here is a troubling quote from Munkres'

We assert that every interval of the form $(x,y]$ is open in $X$: If $X$ has a largest element and $y$ is that element, then $(x,y]$ is just a basis element about $y$. If $y$ is not the largest element of $X$, then $(x,y]$ equals the open set $(x,y')$, where $y'$ is the immediate successor of $y$. 

First, what if $y$ isn't the largest element; will $(x,y]$ still be open? Secondly, what if $y$ has no immediate successor? I don't quite follow Munkres' reasoning.

REPLY [4 votes]: Assume $y$ is not the greatest element in $X$.
$X$ is a well ordered set.  (Which the reals and the rationals are not which makes everything about Munkres' argument counterintuitive.  That is to say, the "usual" meaning of "<" is not a well-ordered one.)
So $S = \{s \in X|s > y\}$ will have a least element.  Call it $y'$.
Then we have $v < y' \iff v \le y$ (which would never be the case if $X$ were the real numbers under the "usual" order.)
So $(x, y] = \{v \in X| x < v \le y\} = \{v \in X|x < v < y'\} = (x,y')$ which is an open set.
If $y$ is the greatest element in $S$. Then $(x,y] = \{v\in X| v > x\}=(x,\infty)$ [if Munkres uses that notation] and that is a basis open set.
Either way $(x,y]$ is an open set.
.....
This is obviously very different than in the rationals or the reals where there would never be any $y$ and $y'$ where $(x , y] = (x,y')$.  However where $X = \mathbb Z$ then $(x ,y] = \{n| x < n \le y\} = \{n|x < n < y+1\} = (x,y+1)$.
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