Corollary 4.7.2.5. Let $(S, \leq )$ be a well-ordered set of order type $\alpha $. Then the cardinality $\kappa = |S|$ is the largest cardinal which satisfies $\kappa \leq \alpha $.

$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$

**Proof.**
The inequality $\kappa \leq \alpha $ follows immediately from the definition of $|S|$. Let $\lambda $ be another cardinal satisfying $\lambda \leq \alpha $. Then $\lambda $ is the order type of an initial segment $S_0 \subseteq S$, so we have $\lambda = | S_0 | \leq | S | = \kappa $.
$\square$