# Why is the smallest element of a well-ordered set in the order topology both open and closed?

In the proof of Theorem 32.4. of Munkres' Topology it is claimed without proof that

Let $X$ be a well-ordered set. ... Assume that the closed set $A$ contains the smallest element $a_0$ of $X$. The set ${\{a_0}\}$ is both open and closed in $X$.

It is closed of course but why open? Even in the simplest example $A=[1,2]$ and $X=[1,3]$, ${\{1}\}$ is not open as $(1,3]$ is not closed!

You have to use the fact that $X$ is well-ordered, not just totally ordered. If $X=\{a_0\}$, then clearly it is open. Otherwise, $X\setminus\{a_0\}$ has a least element $a_1$. The set $\{a_0\}$ is then equal to the open interval $(-\infty, a_1)$.
• I don't understand how $\{a_0\}$ is 'equal' to $(-\infty, a_1)$? – Liebe Oct 9 '16 at 19:09
• If $b<a_1$ and $b\neq a_0$, then $b$ would be a smaller element of $X\setminus\{a_0\}$. – Eric Wofsey Oct 9 '16 at 19:10
• Maybe we can say that $\{a_0\} = (-\infty, a_1) \cap X$ so is open as the intersection of two open set? – Liebe Oct 9 '16 at 19:11
• Whenever you have a totally ordered set $X$ and an element $a\in X$, the notation $(-\infty,a)$ is defined to mean the set $\{x\in X:x<a\}$. (Of course, this is ambiguous if $X$ is contained in some larger totally ordered set, so you should use this notation with care.) – Eric Wofsey Oct 9 '16 at 19:15