Finite Union of Finite Sets is Finite/Proof 2

Proof
Let $S = \set {A_1, \ldots, A_n}$ such that $A_k$ is finite $\forall k = 1, \ldots, n$.

Set:
 * $m = \max \set {\card {A_1}, \ldots, \card {A_n} }$

Then:
 * $\ds \card {\bigcup_{k \mathop = 1}^n A_k} \le \sum_{k \mathop = 1}^n \card {A_k} \le \sum_{k \mathop = 1}^n m = n m$

Hence the result.

Also see

 * Inclusion-Exclusion Principle