Corollary 3

Every subset A of a well-ordered set E is isomorphic to a segment of E.

By virtue of Theorem 3 it is enough to prove that there exists no isomorphism $g$ of E onto a segment of A of the form $S_a$. If there were, $g$ would then be a strictly increasing mapping of E into E such that $g(a)\in S_a$, in other words such that $g(a)<a$; but this inequality contradicts Lemma 4 (with $f$ as the identity mapping).