Well ordered set and both open and closed

Solution 1:

You are misquoting Munkres. He did not say $a$ is the smallest element of just $A$, $a$ was supposed to be the smallest element of the whole set $X$.

Indeed, in the theorem, the part you are talking about is

Finally, assume that $A$ and $B$ are disjoint closed sets in $X$, and $A$ contains the smallest element $a_0$ of $X$. The set $\{a_0\}$ is both open and closed in $X$."

This is in fact true.

• The fact that $\{a_0\}$ is closed in $X$ is trivial: it is just the complement of the open ray $\{x \in X\ |\ a_0 < x\}$.
• $\{a_0\}$ is also open in $X$. This requires slightly more details but isn't too hard. Indeed, if $a_0$ is the largest element of $X$, then we are done. Otherwise, using the well-order on $X$, $a_0$ has an immediate successor $a_1 \in X$. Now notice that the interval $[a_0, a_1)$ is a basis set of the order topology. But this interval is just $\{a_0\}$; so $\{a_0\}$ is open.