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  "body": "\n## Proof\n\nWe count how often each element is counted on the right side, a form of [[double counting|0F8YBHF2W3]].\n\n1. Let $A_1, \\dots, A_n$ be finite sets, and let $U = \\bigcup_{i=1}^{n} A_i$. Then $U$ is the union of the pairwise disjoint sets $A_i \\setminus (A_1 \\cup \\dots \\cup A_{i-1})$. Each is a subset of the finite set $A_i$, so the [[rule of sum|0EE2V3CHFK]] shows that $U$ is finite.\n2. Let $\\mathcal{I}$ be the set of nonempty subsets $I \\subseteq \\set{1, \\dots, n}$. For $I \\in \\mathcal{I}$, let $A_I = \\bigcap_{i \\in I} A_i$. Then $A_I \\subseteq U$.\n3. For $x \\in U$ and $I \\in \\mathcal{I}$, let $e(x, I) = 1$ if $x \\in A_I$, and $e(x, I) = 0$ otherwise.\n4. For each $I \\in \\mathcal{I}$, the set $A_I$ is the disjoint union of the singletons $\\set{x}$ with $x \\in A_I$. So the [[rule of sum|0EE2V3CHFK]] gives $|A_I| = \\sum_{x \\in U} e(x, I)$.\n5. Let $R$ be the right side of the statement. By step 4, and by exchanging two finite sums,\n$$R = \\sum_{I \\in \\mathcal{I}} (-1)^{|I| + 1} \\sum_{x \\in U} e(x, I) = \\sum_{x \\in U} c(x), \\qquad c(x) = \\sum_{I \\in \\mathcal{I}} (-1)^{|I| + 1} e(x, I).$$\n6. Fix $x \\in U$. Let $J = \\set{i : x \\in A_i}$ and $m = |J|$. Since $x \\in U$, $J$ is nonempty, so $m \\ge 1$.\n7. For $I \\in \\mathcal{I}$, we have $x \\in A_I$ exactly when $x \\in A_i$ for all $i \\in I$, that is, when $I \\subseteq J$.\n8. So $c(x)$ is the sum of $(-1)^{|I| + 1}$ over the nonempty subsets $I$ of $J$.\n9. By the [[number of $k$-element subsets|0BFMR2FQ8B]], $J$ has $\\binom{m}{k}$ subsets with $k$ elements. Grouping the subsets by size gives\n$$c(x) = \\sum_{k=1}^{m} (-1)^{k+1} \\binom{m}{k}.$$\n10. Since $m \\ge 1$, the [[alternating sum of binomial coefficients|0PQQGFAWSH]] gives $\\sum_{k=0}^{m} (-1)^k \\binom{m}{k} = 0$.\n11. The term with $k = 0$ there is $\\binom{m}{0} = 1$. So $\\sum_{k=1}^{m} (-1)^k \\binom{m}{k} = -1$, and by step 9, $c(x) = 1$.\n12. By steps 5 and 11, $R = \\sum_{x \\in U} 1 = |U|$. This is the claimed formula $\\left| \\bigcup_{i=1}^{n} A_i \\right| = \\sum_{I \\in \\mathcal{I}} (-1)^{|I| + 1} |A_I|$.\n\n## Idea\n\nExpand each $|A_I|$ as a sum over its elements and collect the contributions of one element $x$. If $x$ lies in exactly $m \\ge 1$ of the sets, the signs add up to $\\binom{m}{1} - \\binom{m}{2} + \\dots$, which is $1$ because $(1 - 1)^m = 0$. So every element of the union is counted exactly once.\n",
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