Union of Finite Sets is Finite/Proof 1

Proof
If $S$ or $T$ is empty, the result is trivial.

Otherwise, let $f: \N_{<n} \to S$ and $g: \N_{<m} \to T$ be bijections, where $\N_{<n}$ is an initial segment of $\N$.

Now define $h: \N_{< n+m} \to S \cup T$ by:


 * $h (i) = \begin{cases}

f (i) & \text{if $i < n$} \\ g (i - n) & \text{if $i \ge n$} \end{cases}$

By Set Finite iff Surjection from Initial Segment of Natural Numbers, it suffices to show that $h$ is surjective.

Suppose that $s \in S$. Then:

Next suppose that $t \in T$.

Note that $n + g^{-1}(t) < n + m$, so that $n + g^{-1}(t) \in \N_{< n+m}$.

Then:

Hence $h$ is a surjection.

Therefore $S \cup T$ is finite.