A Note on Intervals in the Hales-Jewett Theorem

The Hales-Jewett theorem for alphabet of size 3 states that whenever the Hales-Jewett cube [3]^n is r-coloured there is a monochromatic line (for n large). Conlon and Kamcev conjectured that, for any n, there is a 2-colouring of [3]^n for which there is no monochromatic line whose active coordinate set is an interval. In this note we disprove this conjecture.


Introduction
In order to state the Hales-Jewett theorem we need some notation. Given positive integers k and n let [k] n be the set of all words in symbols {1, . . . , k} of length n. A set L ⊂ [k] n is called a combinatorial line if there exist a nonempty set S ⊂ [n] and integers a i ∈ [k] for all i ∈ S such that L = {(x 1 , . . . , x n ) : x i = a i for all i ∈ S, x i = x j for all i, j ∈ S}. The set S is called the active coordinate set of L.
Theorem (Hales-Jewett [2]). For any k and r there exists n such that whenever [k] n is r-coloured there is a monochromatic combinatorial line.
As noted by Conlon and Kamcev in [1], by following Shelah's proof [3] of the Hales-Jewett theorem it can be shown that one can always find a monochromatic combinatorial line whose active coordinate set S is an union of at most HJ(k − 1, r) intervals, where HJ(k − 1, r) is the smallest integer n for which the Hales-Jewett theorem holds for k − 1 and r.
In the case k = 3, since HJ(2, r) = r, this says that one can always find a monochromatic line whose active coordinate set is a union of at most r intervals. Conlon and Kamcev proved in [1] that this bound is tight for r odd: in other words, they showed that for each odd r there is an r-colouring of [3] n (for any n) for which every monochromatic line has active coordinate set made up of at least r intervals. They conjectured that this would also be the case for r even. In particular, for r = 2, they conjectured that for all n there exists a 2-colouring of [3] n for which there exists no monochromatic combinatorial line whose active coordinate set is an interval.
In this note we will prove that, perhaps surprisingly, their conjecture is false. This can be stated in the following form.
Theorem 1. For all sufficiently large n, whenever [3] n is 2-coloured there exists a monochromatic combinatorial line whose active coordinate set is an interval.

The proof of Theorem 1
The idea of the proof is as follows. By applying Ramsey's theorem, we will pass to a subspace on which the colour of a word depends only on its 'pattern' of intervals. Once this is done, we can consider some particular small patterns.
For a word w let w be obtained from w by contracting every interval on which w is a constant to a single letter. We will consider the particular words s 1 = 132 , s 2 = 1232, s 3 = 1312, s 4 = 13232 and s 5 = 13132. Set t i to be the length of the word s i . Put n 0 = 4 and for 1 ≤ i ≤ 5 let n i = R (ti−1) (n i−1 ) where R (t) (s) = R (t) (s, s) is the t-set Ramsey-number. Finally set n = n 5 + 1.
Let c be any 2-colouring of [3] n . For a word w we define the set of breakpoints is monochromatic for the colouring c i , say with colour d i .
Thus we obtain sets T 0 ⊂ T 1 ⊂ . . . ⊂ T 5 with |T 0 | ≥ 4 and colours d 1 . . . , d 5 such that c i restricted to T (ti+1−1) i is constant with value d i+1 . Note that it is impossible to choose colours d 1 , . . . , d 5 without at least one of the following sets having just one element (i.e. all colours being equal). Indeed, if all the sets N 1 , N 2 , N 3 , N 4 contain both colours we must have d 2 = d 3 and d 1 = d 4 = d 5 , which implies that |N 5 | = 1.
Let a 1 < a 2 < a 3 < a 4 be elements of T 0 . We will use the shorthand for the word which has w i = b j for all a j−1 < i ≤ a j with the convention a 0 = 0 and a 5 = n. Note that we will allow b i = b i+1 . Hence It is straightforward to verify that It is clear that the colours used to colour elements of the line L i are exactly the colours in the set N i . As observed earlier, one of the sets N i contains only one colour, which implies that the associated line L i is monochromatic. Since all the sets S i are intervals, this completes the proof.