Kerodon

Proof. Choose well-orderings $(S, \leq _ S)$ and $(T, \leq _ T)$ having order types $|S|$ and $|T|$, respectively. If $|S| \leq |T|$, then there is an isomorphism of $(S, \leq _{S} )$ with an initial segment of $(T, \leq _{T} )$; this isomorphism in particular gives a monomorphism of sets $S \hookrightarrow T$. For the converse, suppose that there exists a monomorphism $f: S \hookrightarrow T$. Then there is a unique linear ordering $\leq '_{S}$ on the set $S$ for which $f$ defines a strictly increasing function $(S, \leq '_{S} ) \rightarrow (T, \leq _{T} )$. Then $\leq '_{S}$ is a well-ordering (Remark 4.7.1.5); let $\alpha $ denote its order type. We then have $|S| \leq \alpha \leq |T|$, where the second inequality follows from Remark 4.7.1.21. $\square$