Finite Union of Finite Sets is Finite

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Theorem

Let $S$ be a finite set of finite sets.


Then the union of $S$ is finite.


Proof 1

The proof proceeds by induction.

Let $S$ be a finite set with cardinality $n$.

If $n = 0$ then $S = \O$, so $\bigcup S = \O$, which is finite.

Suppose that the union of any finite set of finite sets with cardinality $n$ has a finite union.



Let $S$ have cardinality $n^+$.

Then there is a bijection $f: n^+ \to S$.

Then:

$\ds \bigcup S = \bigcup_{k \mathop \in n^+} \map f k = \bigcup_{k \mathop \in n} \map f k \cup \map f n$


By Union of Finite Sets is Finite, $\bigcup S$ is finite.

$\blacksquare$


Proof 2

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.

$\blacksquare$


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