Finite Sets are Comparable

Theorem
Let $S$ and $T$ be finite sets.

Then $S$ and $T$ are comparable by size.

Proof
By definition of finite set, there exist $m, n \in \N$ such that:


 * $S \sim \N_{<n}$
 * $T \sim \N_{<m}$

That is, there are bijections $f, g$:


 * $f: S \to \N_{<n}$
 * $g: T \to \N_{<m}$

, suppose that $m \le n$.

Then $\N_{<m} \subseteq \N_{<n}$, and so we can define:


 * $h: T \to S, \map h t = \map {f^{-1}} { \map g t }$

where $f^{-1}: \N_{<n} \to S$ is the inverse of $f$.

By Inverse of Bijection is Bijection and Composite of Injections is Injection, it follows that $h$ is injective.

By Injection to Image is Bijection, $h$ is a bijection to $\Img h$.

Since $\Img h \subseteq S$, it follows that $T$ is comparable to $S$.