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  "body": "\n## Proof\n\nWe first show that every subset of a finite set is finite, by induction on the size. Then we apply the [[rule of sum|0EE2V3CHFK]].\n\n1. Claim: for each natural number $n$, every subset of a set with $n$ elements is finite. We argue by induction on $n$.\n2. For $n = 0$, the only set with $0$ elements is $\\varnothing$. Its only subset is $\\varnothing$, which is finite.\n3. Suppose the claim holds for $n$. Let $V$ have $n + 1$ elements, with a [[bijection|0SNQEF7Z53]] $g : \\set{1, \\dots, n+1} \\to V$. Let $B \\subseteq V$.\n4. Put $v = g(n+1)$ and $V' = V \\setminus \\set{v}$. The restriction of $g$ to $\\set{1, \\dots, n}$ is a bijection onto $V'$, so $V'$ has $n$ elements.\n5. The set $B \\setminus \\set{v}$ is a subset of $V'$. By the induction hypothesis it is finite.\n6. If $v \\notin B$, then $B = B \\setminus \\set{v}$ is finite.\n7. If $v \\in B$, then $B$ is the union of the [[disjoint|0W4RPA8QH3]] finite sets $B \\setminus \\set{v}$ and $\\set{v}$. By the rule of sum, $B$ is finite. This proves the claim.\n8. Now let $U$ be finite and $A \\subseteq U$. By the claim, $A$ and $A^c = U \\setminus A$ are finite.\n9. Every element of $U$ is in $A$ or in $A^c$, and no element is in both. So $U$ is the disjoint union of $A$ and $A^c$.\n10. By the rule of sum, $|U| = |A| + |A^c|$. Hence $|A^c| = |U| - |A|$.\n\n## Idea\n\nThe set $U$ splits into the part inside $A$ and the part outside $A$, with no overlap.\nSo the two sizes add up to $|U|$, and the size of the outside part is a difference.\nThe only extra work is to check that both parts are finite.\n",
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