Q: Let $≤$ be a total order on $A$. For each $u\in A$, let $s(u)=\{x\in A:x<u\}$. Suppose that for every $u\in A$ and $X$, if $X\subseteq s(u)$ is nonempty, then $X$ has a $≤$-least element. Prove that $≤$ is a well-ordering on $A$.
A: I understand that, since $≤$ is a total order on $A$, we need only show that any nonempty subset of $A$ has a least element. I'm sort of seeing that $X$ is these subsets which we're supposing has a least element? But it would then be true by the supposition, but that can't be. Any help is appreciated.
Suppose $B \subseteq A$ is non-empty; we need to show it has a minimum. Pick $b \in B$. If $s(b) \cap B$ is empty this means that there are no elements of $B$ that are strictly smaller than $b$. As $\le$ is a total order, this means that $b=\min(B)$.
So if $s(b) \cap B$ is empty we are done. Otherwise, we apply the assumption that any non-empty subset of $s(b)$ has a $\le$-least element, so we get some $b_0$ that is $\le$-minimal in $s(b) \cap B$.
Now if $x \in B$ is arbitary, either $b_0 \le x$ or $x \le b_0$ must hold. If $x \le b_0 (< b)$, then $x \in s(b) \cap B$ and $\le$-minimality tells us that $b_0=x$ and we're done. So $b_0 \le x$ and $b_0$ is also minimal for $B$.