Van Der Waerden's Theorem - Proof

Proof

A D-dimensional arithmetic progression consists of numbers of the form:

where a is the basepoint, the s's are the different step-sizes, and the i's range from 0 to L-1. A d-dimensional AP is homogenous for some coloring when it is all the same color.

A D-dimensional arithmetic progression with benefits is all numbers of the form above, but where you add on some of the "boundary" of the arithmetic progression, i.e. some of the indices i's can be equal to L. The sides you tack on are ones where the first k i's are equal to L, and the remaining i's are less than L.

The boundaries of a D-dimensional AP with benefits are these additional arithmetic progressions of dimension d-1,d-2,d-3,d-4, down to 0. The 0 dimensional arithmetic progression is the single point at index value (L,L,L,L...,L). A D-dimensional AP with benefits is homogenous when each of the boundaries are individually homogenous, but different boundaries do not have to necessarily have the same color.

Next define the quantity MinN(L, D, N) to be the least integer so that any assignment of N colors to an interval of length MinN or more necessarily contains a homogenous D-dimensional arithmetical progression with benefits.

The goal is to bound the size of MinN. Note that MinN(L,1,N) is an upper bound for Van-Der-Waerden's number. There are two inductions steps, as follows:

1. Assume MinN is known for a given lengths L for all dimensions of arithmetic progressions with benefits up to D. This formula gives a bound on MinN when you increase the dimension to D+1:

let

Proof: First, if you have an n-coloring of the interval 1...I, you can define a block coloring of k-size blocks. Just consider each sequence of k colors in each k block to define a unique color. Call this k-blocking an n-coloring. k-blocking an n coloring of length l produces an n^k coloring of length l/k.

So given a n-coloring of an interval I of size M*MinN(L,1,n^M)) you can M-block it into an n^M coloring of length MinN(L,1,n^M). But that means, by the definition of MinN, that you can find a 1-dimensional arithmetic sequence (with benefits) of length L in the block coloring, which is a sequence of blocks equally spaced, which are all the same block-color, i.e. you have a bunch of blocks of length M in the original sequence, which are equally spaced, which have exactly the same sequence of colors inside.

Now, by the definition of M, you can find a d-dimensional arithmetic sequence with benefits in any one of these blocks, and since all of the blocks have the same sequence of colors, the same d-dimensional AP with benefits appears in all of the blocks, just by translating it from block to block. This is the definition of a d+1 dimensional arithmetic progression, so you have a homogenous d+1 dimensional AP. The new stride parameter s_{D+1} is defined to be the distance between the blocks.

But you need benefits. The boundaries you get now are all old boundaries, plus their translations into identically colored blocks, because i_{D+1} is always less than L. The only boundary which is not like this is the 0 dimensional point when . This is a single point, and is automatically homogenous.

2. Assume MinN is known for one value of L and all possible dimensions D. Then you can bound MinN for length L+1.

proof: Given an n-coloring of an interval of size MinN(L,n,n), by definition, you can find an arithmetic sequence with benefits of dimension n of length L. But now, the number of "benefit" boundaries is equal to the number of colors, so one of the homogenous boundaries, say of dimension k, has to have the same color as another one of the homogenous benefit boundaries, say the one of dimension p

if

has the same color as

then

have the same color

i.e. u makes a sequence of length L+1.

This constructs a sequence of dimension 1, and the "benefits" are automatic, just add on another point of whatever color. To include this boundary point, one has to make the interval longer by the maximum possible value of the stride, which is certainly less than the interval size. So doubling the interval size will definitely work, and this is the reason for the factor of two. This completes the induction on L.

Base case: MinN(1,d,n)=1, i.e. if you want a length 1 homogenous d-dimensional arithmetic sequence, with or without benefits, you have nothing to do. So this forms the base of the induction. The VanDerWaerden theorem itself is the assertion that MinN(L,1,N) is finite, and it follows from the base case and the induction steps.