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PROBABILISTIC METHODS IN COMBINATORICS 1. Introduction ...

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<strong>PROBABILISTIC</strong> <strong>METHODS</strong> <strong>IN</strong> COMB<strong>IN</strong>ATORICS 3<br />

The events E R and E S are independent when the subgraphs induced by R and S do not share<br />

edges. Thus is E R and E S are dependent, then we necessarily have |R ∩ S| ≥ 2. For a fixed R,<br />

there are at most ( )(<br />

k n<br />

2 k−2)<br />

choices of S with |S| = k and |R ∩ S| ≥ 2.<br />

Applying the Lovász local lemma to the events {E R : R ⊂ V (K n ), |R| = k} and p = 2 1−(k 2) and<br />

d = ( )(<br />

k n<br />

2 k−2)<br />

, we see that with positive probability none of the events ER occur, thereby giving a<br />

colouring with no monochromatic K k ’s.<br />

□<br />

3. Set systems<br />

In this section we apply the probabilistic method to two extremal problems regarding families of<br />

subsets of {1, 2, . . . , n}.<br />

3.<strong>1.</strong> Antichains. Let F be a collection of subsets of {1, 2, . . . , n} such that no set in F is contained<br />

in another set in F. Such a collection is called an antichain. We would like to know what is the<br />

maximum number of sets in an antichain.<br />

If F is the collection of all size k subsets of {1, 2, . . . , n}, then F is an antichain, since no k-<br />

element set can contain another k-element set. This gives |F| = ( n<br />

k)<br />

, which is maximized when<br />

k = ⌊ n<br />

2<br />

⌋<br />

or<br />

⌈ n<br />

2<br />

⌉<br />

. The next result shows that we cannot do better.<br />

Theorem 3.1 (Sperner). If F is an antichain of subsets of {1, 2, . . . , n}, then |F| ≤<br />

( ) n<br />

.<br />

⌊n/2⌋<br />

Proof. Consider a random permutation σ of {1, 2, . . . , n}, and its associated chain of subsets<br />

∅, {σ(1)} , {σ(1), σ(2)} , {σ(1), σ(2), σ(3)} , . . . , {σ(1), . . . , σ(n)}<br />

where the last set is always equal to {1, 2, . . . , n}. For each A ⊂ {1, 2, . . . , n}, let E A denote the<br />

event that A is found in this chain. Then<br />

|A|!(n − |A|)!<br />

Pr(E A ) = = 1 1<br />

(<br />

n!<br />

n<br />

) ≥ ( n<br />

).<br />

|A| ⌊n/2⌋<br />

Indeed, if A were to be found in the chain, it must occur in the (|A| + 1)-th position, and are |A|!<br />

ways to form a chain in front of it, and (n − |A|)! ways to continue the chain after it.<br />

Since F is an antichain, if A, B ∈ F are distinct, then E A and E B cannot both occur. Thus<br />

1<br />

{E A : A ∈ F} is a set of disjoint events, each with probability at least<br />

( ⌊n/2⌋) . Thus |F| ≤ ( n<br />

⌊n/2⌋)<br />

.<br />

n<br />

3.2. Intersecting family. We say that family F of sets is intersecting if A ∩ B ≠ ∅ for every<br />

A, B ∈ F. Let F be an intersecting family of k-element subsets of {1, 2, . . . , n}. How large can |F|<br />

be?<br />

If we take F to be the collection of all k-element subsets of {1, 2, . . . , n} containing the element<br />

1, then F is intersecting, and |F| ≤ ( n−1<br />

k−1)<br />

. Now we show that this is the best we can do.<br />

Theorem 3.2 ( (Erdős-Ko-Rado). )<br />

If F is a intersecting family of k-element subsets of {1, 2, . . . , n},<br />

n − 1<br />

then |F| ≤ .<br />

k − 1<br />

Proof. Arrange 1, 2, . . . , n randomly around a circle. For each k-element subset of A of {1, 2, . . . , n},<br />

we say that A is contiguous if all the elements of A lie in a contiguous block on the circle. The<br />

probability that A forms a contiguous set on the circle is exactly<br />

( n , since there are n possible<br />

n<br />

k)<br />

positions for a continuous block of length k on the circle, and for each such choice, the probability<br />

1<br />

that A falls exactly into those position is . It follows that the expected number of continguous<br />

( n k)<br />

sets in F is exactly n|F|<br />

( n k) .<br />

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