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  "body": "\n## Proof\n\nWe argue by contradiction.\n\n1. Let $A$ be an [[infinite set|01MKW4R283]], let $B$ be a [[finite set|0BBTH3YSZA]], and let $f : A \\to B$ be a function.\n2. For $b \\in B$, write $A_b = f^{-1}(b) = \\set{a \\in A : f(a) = b}$.\n3. Suppose, for a contradiction, that $A_b$ is finite for every $b \\in B$.\n4. The sets $A_b$ for $b \\in B$ are pairwise disjoint. Indeed, if $a \\in A_b \\cap A_{b'}$, then $b = f(a) = b'$.\n5. Their union is $A$, because each $a \\in A$ lies in $A_{f(a)}$.\n6. Since $B$ is finite, $A$ is the union of finitely many pairwise disjoint finite sets. If $B$ is empty, this union is empty.\n7. By the [[rule of sum|0EE2V3CHFK]], such a union is finite. So $A$ is finite.\n8. This contradicts step 1, since an infinite set is by definition not finite.\n9. Hence some $b \\in B$ has an infinite fibre $f^{-1}(b)$.\n\n## Idea\n\nThe fibres of $f$ cover $A$, and there are only finitely many of them. A finite union of finite sets is finite, so one fibre must be infinite.\n",
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