In a well ordered set $X$, a subset $S$ is an initial segment if
$S = \{x\in X: x\leq a\}$ for some $a\in X$ or $S = X$
Give $x\in X$, and $s\in S$, if $x\leq s$ then $x\in S$
I can see why $1$ implies $2$, but how can I proof $2$ implies $1$? How do I show the existence of $a\in X$, I think in some sense, $a$ is a "maximal element", but well-ordered sets only give the existence of the smallest element.
A side question, we know the set $X-S$ has a smallest element $y\in X-S$, but we don't have $S= \{x\in X : x\leq y\}$, because this would imply $y\in S$ as well since $y\leq y$, this seemed a little counter intuitive for me...
Thank you!