Proof of the infinite pigeonhole principle

Proof

We argue by contradiction.

1. Let $A$ be an infinite set, let $B$ be a finite set, and let $f : A \to B$ be a function.
2. For $b \in B$, write $A_b = f^{-1}(b) = \set{a \in A : f(a) = b}$.
3. Suppose, for a contradiction, that $A_b$ is finite for every $b \in B$.
4. The sets $A_b$ for $b \in B$ are pairwise disjoint. Indeed, if $a \in A_b \cap A_{b'}$, then $b = f(a) = b'$.
5. Their union is $A$, because each $a \in A$ lies in $A_{f(a)}$.
6. Since $B$ is finite, $A$ is the union of finitely many pairwise disjoint finite sets. If $B$ is empty, this union is empty.
7. By the rule of sum, such a union is finite. So $A$ is finite.
8. This contradicts step 1, since an infinite set is by definition not finite.
9. Hence some $b \in B$ has an infinite fibre $f^{-1}(b)$.