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  "body": "\n## Statement\n\nLet $n$ and $k$ be positive integers with $n \\ge 2k$.\nLet $\\mathcal{F}$ be a [[family|0C21Q694AG]] of $k$-element subsets of $\\set{1, \\ldots, n}$ that is intersecting: $A \\cap B \\ne \\emptyset$ for all $A, B \\in \\mathcal{F}$.\nThen\n$$|\\mathcal{F}| \\le \\binom{n - 1}{k - 1}.$$\n\n## Remarks\n\nKatona's proof places $1, \\ldots, n$ around a circle in a random order.\nAn intersecting family contains at most $k$ of the $n$ arcs of $k$ consecutive points, when $n \\ge 2k$.\nEach $k$-set is such an arc with probability $n / \\binom{n}{k}$, so averaging gives $|\\mathcal{F}| \\cdot n / \\binom{n}{k} \\le k$, that is, $|\\mathcal{F}| \\le \\frac{k}{n} \\binom{n}{k} = \\binom{n-1}{k-1}$.\n\nExample: the star of all $k$-sets that contain the element $1$ is intersecting and has exactly $\\binom{n-1}{k-1}$ members.\nFor $n \u003e 2k$ the stars are the only families that attain the bound.\nFor $n = 2k$ there are others, for example one set from each complementary pair.\nThe hypothesis $n \\ge 2k$ is necessary: if $k \u003c n \u003c 2k$, any two $k$-subsets intersect, and there are $\\binom{n}{k} = \\frac{n}{k} \\binom{n-1}{k-1} \u003e \\binom{n-1}{k-1}$ of them.\n\n## Explanations\n\n### L1 human\n\nSuppose every two committees of $k$ people, chosen from $n$ people, must share a member.\nHow many committees can there be?\nThe simple answer is to give every committee one fixed common member.\nThe theorem says that you cannot do better, provided the group is at least twice the committee size.\n\n### L2 human\n\nThe theorem is a basic result of extremal set theory, with many extensions: to $t$-intersecting families, to vector spaces, to permutations, and to other structures.\nIts circle proof is an averaging argument of the same kind as the proof of the [[LYM inequality|D1GQNNRDDN]].\nThe [[Kruskal–Katona theorem|G6WNYB0WMY]] gives another proof.\nIn graph language it gives the independence number of the Kneser graph.\n",
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