Mancosu - Philosophy of Mathematical Practice (Oxford, 2008).pdf

understanding proofs 337value of k, thereisann large enough, so that no matter how one colors thethe complete graph of n vertices edges red and blue, there is a homogeneoussubset of size k.This raises the difficult problem of determining, for a given value of k, howlarge n has to be. To show that, for a given k, avalueofn is not large enoughmeans showing that there is a graph of size n with no homogeneous subsetof size k. PaulErdös pioneered a method, called the probabilistic method, forproviding such lower bounds: one imagines that a coloring is chosen at randomfrom among all colorings of the complete graph on n vertices, and then oneshows that with nonzero probability, the graph will have no homogeneoussubset of size k.Theorem 1. For all k ≥ 2, if n < 2 k/2 , there is a coloring of the complete graph on nvertices with no homogeneous subset of size k.Proof. For k = 2 this is trivial, and for k = 3 this is easily verified by hand. Sowe can assume k ≥ 4.Suppose n < 2 k/2 , and consider all red–blue colorings, where we color eachedge independently red or blue with probability 1/2. Thus all colorings areequally likely with probability 2 −( n 2 ) .LetA be a set of vertices of size k. Theprobability of the event A R that the edges in A are all colored red is then 2 −( k 2 ) .Hence it follows that the probability p R for some k-set to be colored all red isbounded byp R = Prob ⋃A R ≤ ∑( ) nProb(A R ) = 2 −( k 2 ) .k|A|=k|A|=kNow for k ≥ 2, we have ( n k ) = n(n−1)(n−2)···(n−k+1) ≤ nk.Soforn < 2 k/2 andk(k−1)···12 k−1k ≥ 4, we have( ) n2 −( k 2 ) ≤ nk kk 2 k−1 2−( 2 ) < 2 k22 −( k 2 )−k+1 = 2 − k 2 +1 ≤ 1/2.Since p R < 1/2, and by symmetry p B < 1/2 for the probability of some k verticeswith all edges between them colored blue, we conclude that p R + p B < 1forn < 2 k 2 ,sotheremust be a coloring with no red or blue homogeneous subset ofsize k.□The text of this proof has been reproduced with only minor modificationsfrom Aigner and Ziegler’s Proofs from the Book (2001). (For sharper boundsand more information see Graham et al., 1994.) To make sense of the proof,remember that ( n k ) = n! is the number of ways of choosing a subset ofk!(n−k)!k elements from a set of n objects. The details of the argument are not soimportant; I am specifically interested in the chain of inequalities. You may