Erdős–Ko–Rado theorem

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In combinatorial mathematics, the Erdős–Ko–Rado theorem of Paul Erdős, Chao Ko, and Richard Rado is the following. If $n\geq2r$ , and $A$ is a family of subsets of ${1,2,..., n}$ , such that each subset is of size $r$ , each pair of subsets intersects, and A is an antichain (there are no two groups $A_1,A_2 \in A$ such that $A_1 \subset A_2$ ), then the maximum number of sets that can be in $A$ is given by the binomial coefficient

${n-1} \choose {r-1}$ .

According to Erdős the theorem was proven in 1938, but was not published until 1961 in an apparently more general form. The subsets in question were only required to be size at most $r$ , and with the additional requirement that no subset be contained in any other. This statement is not actually more general: any subset that has size less than $r$ can be increased to size $r$ to make the above statement apply.

[edit] Proof

The original proof of 1961 used induction on $n$ . In 1972, Gyula O. H. Katona gave the following short and beautiful proof using double counting.

Suppose we have some such set $A$ . Arrange the elements of ${1,2,..., n}$ in any cyclic order, and inquire how many sets from $A$ can form intervals within this cyclic order. For example if $n = 8$ and $r = 3$ , we could arrange the numbers $1,...,8$ as

[3,1,5,4,2,7,6,8]

and intervals would be

{1,3,5},{1,4,5},{2,4,5},{2,4,7},{2,6,7},{6,7,8},{3,6,8},{1,3,8}

(Key step) At most $r$ of these intervals can be in $A$ . If

(a 1, a 2,..., a r)

is one of these intervals in $A$ then for every $i$ , there is at most one interval in $A$ which separates $a i$ from $a i + 1$ , i.e. contains precisely one of $a i$ and $a i + 1$ . (If there were two, they would be disjoint, since $n\geq2r$ .) Furthermore, if there are $r$ intervals in $A$ , then they must contain some element in common.

There are $(n - 1)!$ cyclic orders, and each set from $A$ is an interval in precisely $r!(n - r)!$ of them. Therefore the average number of intervals that $A$ has in a random cyclic order must be

$\frac{|A|\cdot r!(n-r)!}{(n-1)!}\leq r.$

Rearranging the inequality, we get

$|A|\leq\frac{r(n-1)!}{r!(n-r)!}=\frac{(n-1)!}{(r-1)!(n-r)!}={n-1\choose r-1},$

establishing the theorem.

[edit] Further reading

P. Erdős. My joint work with Richard Rado. Surveys in Combinatorics, London Math. Soc. Lecture Note Series 123 (1987), pages 53-80.

P. Erdős, C. Ko, R. Rado. Intersection theorems for systems of finite sets, Quarterly Journal of Mathematics, Oxford Series, series 2, volume 12 (1961), pages 313--320.