# Union of Finite Sets is Finite

## Theorem

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

Then $S \cup T$ is a finite set.

## Proof 1

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:

$\map h i = \begin{cases} \map f i : & \text {if$i < n$} \\ \map g {i - n} : & \text{if$i \ge n$} \end{cases}$

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

Let $s \in S$.

Then: