By Stefan H. M. van Zwam

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**Additional info for Combinatorial Mathematics [Lecture notes]**

**Sample text**

Then | | ≤ n. If | | = n, say = {A1 , . . , An }, then one of the following holds: (i) Up to relabeling, Ai = {i, n} (so An = {n}); (ii) Up to relabeling, Ai = {i, n} for i ∈ [n − 1], and An = [n − 1]; (iii) There exists an integer q > 0 such that n = q2 + q + 1. Each Ai has q + 1 elements, and each element is in q + 1 of the Ai . This result is usually framed in terms of incidence geometry. We think of [n] as a set of points, and of as a set of lines through these points. The lines obey the classical rules of projective geometry: every two lines intersect in exactly one point.

I) If P has a chain of size r, then P cannot be partitioned into fewer than r antichains; 50 EXTREMAL COMBINATORICS (ii) If P has an antichain of size r, then P cannot be partitioned into fewer than r chains. We will prove that the bound of r is tight in both cases. 4 THEOREM. Suppose the longest chain in a poset (P, ≤) has size r. Then we can partition P into r antichains. Proof: For x ∈ P, define the height of x as the longest chain ending in x. Let Ai be the set of elements of height i, for i ∈ [r].

Am } is a two-distance set with distances c, d. Define, for i ∈ [m], f i (x) := x − ai 2 − c2 x − ai 2 − d2 . 1 the f i are linearly independent. To bound m, then, it suffices to find a low-dimensional space containing all of the f i . If we look at the expansion, we see that each f i is a linear combination of n i=1 x i2 2 , n i=1 x i2 x j , xi x j, , xi, , 1. Hence f1 , . .

### Combinatorial Mathematics [Lecture notes] by Stefan H. M. van Zwam

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