Well-ordered system.

Question

I would like to ask you the following question. Let $(S,<)$ is ordered system satisfying the condition: $\forall A\subset S:[\forall a\in S:S_a\subset A \implies a\in A] \implies A=S,$ where $S_a =\{x:x\in S, x<a\}.$ How to prove that $(S,<)$ is well-ordered system? This problem has its origin in the book "The number systems (S. Feferman)" and it is a conversion of the theorem 4.49 remarked without proof . Thank you.

Answer 1

Let $B\subseteq S$ arbitrary nonempty, let $x<B$ denote ($\forall b\in B: x<b$), and set $A:=\{x\in S:x<B\}$.
$A$ is clearly disjoint to $B$, so $A\ne S$, which implies $\exists a\in S: S_a\subseteq A$ but $a\notin A$.
For such an $a$, using linearity of the order, we get $x<a\iff x\in A \iff x<B$, and thus $a=\inf B$ follows.
(Use linearity again, and that $\forall b:x\le b$ means either $x=\min B$ or $x<B$.)
Now if $a\notin B$, then $a$ is strictly smaller than every $b\in B$, contradicting to $a\notin A$.
This proves that $B$ has smallest element, $a$.