Remark 4.7.1.21. Let $(S,\leq )$ and $(T, \leq )$ be well-ordered sets. The following conditions are equivalent:
- $(1)$
There exists an initial segment embedding $f: S \hookrightarrow T$.
- $(2)$
There exists a strictly increasing function $f: S \hookrightarrow T$.
The implication $(1) \Rightarrow (2)$ is immediate from the definitions. To prove the converse, let $f: S \hookrightarrow T$ be a strictly increasing function, and suppose that there is no initial segment embedding from $S$ to $T$. Invoking Corollary 4.7.1.17, we deduce that there is an initial segment embedding $g: T \hookrightarrow S$. The composition $(g \circ f): S \hookrightarrow S$ is strictly increasing, and therefore satisfies $(g \circ f)(s) \geq s$ for each $s \in S$ (Proposition 4.7.1.14). Since the image of $g$ is an initial segment $S_0 \subseteq S$, we must have $S_0 = S$. It follows that $g^{-1}: S \xrightarrow {\sim } T$ is an isomorphism of linearly ordered sets, contradicting our assumption.