TITLE: Prove that every totally ordered set has a well-ordered cofinal subset
QUESTION [10 upvotes]: I thought this was easy until I realized that for a set to be well-ordered, EACH of it's non-empty subsets must have a least element.
Let $A$ be a non-empty totally ordered set. Let $x_0$ be some element of $A$. Define $S=\{x\in A|x\ge x_0\}$. $x_0 \in S$ so $S$ isn't empty.
Now, this seems like it would work. It certainly is totally ordered and cofinal in $A$, but it might not be well-ordered.
For example, if $A=\mathbb R$ then $(x_0 + 1, x_0 + 2)$ (where $()$ denotes an open interval) is a non-empty subset of $S$, but doesn't have a least element.

REPLY [5 votes]: By transfinite recursion, we can define a sequence $(a_\alpha)_{\alpha\in\mathrm{Ord}}$ of elements of $A$ (see ordinals) such that $a_\alpha$ satisfies
$$\forall\beta,\ \beta<\alpha\implies a_\beta<a_\alpha$$
Each step is possible if and only if $(a_\beta)_{\beta<\alpha}$ isn't cofinal.
But this construction must stop somewhere for there isn't an unlimited number of elements of $A$. That's the point where you get a cofinal sequence, which is well ordered because ordinals are.