# 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 $S = \varnothing$, so $\bigcup S = \varnothing$, 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:

- $\displaystyle \bigcup S = \bigcup_{k \mathop \in n^+} f \left({k}\right) = \bigcup_{k \mathop \in n} f \left({k}\right) \cup f \left({n}\right)$

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

$\blacksquare$

## Proof 2

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

Set:

- $m = \max \left\{{ \left|{A_1}\right|, \ldots, \left|{A_n}\right|}\right\}$

Then:

- $\displaystyle \left|{ \bigcup_{k \mathop = 1}^n A_k}\right| \le \sum_{k \mathop = 1}^n \left|{A_k}\right| \le \sum_{k \mathop = 1}^n m = n m$

Hence the result.

$\blacksquare$

## Sources

- 2010: Raymond M. Smullyan and Melvin Fitting:
*Set Theory and the Continuum Problem*(revised ed.) ... (previous) ... (next): Chapter $3$: The Natural Numbers: $\S 6$ Finite Sets: Exercise $6.3$