CEU “Topics in Combinatorics” course
Topics covered in the March 3, 2017 class
Notation: [n] denotes the positive integers up to n, i.e., [n] = {1, 2, . . . , n}. An r-coloring of a
set H is a function f : H → [r]. We call f (x) the color of x. We call a subset A ⊆ H monochromatic
if f (x) = f (y) whenever x, y ∈ A.
A k-term arithmetic progression is a set {a, a + b, a + 2b, . . . , a + (k − 1)b}, where the increment
b is not zero. We write [k]d for the d-dimensional grid of side k, that is [k]d = {(x1 , . . . , xd ) | ∀i ∈
[d] : xi ∈ [k]}. A combinatorial line in [k]d is obtained by fixing some coordinates and making the
remaining coordinates equal. It is OK not to fix any coordinates, but it is not OK to fix all of them.
In other words a combinatorial line is the image of an injective function f : [k] → [k]d satisfying
that each coordinate i 7→ (f (i))j is either a constant or the identity function. An example: the
following three points form a combinatorial line in [3]8 :
(1, 1, 2, 1, 1, 3, 1, 1)
(1, 2, 2, 1, 2, 3, 2, 2)
(1, 3, 2, 1, 3, 3, 3, 3)
Note that each combinatorial line in [k]d contains k elements.
In the following theorems k, r and n denote positive integers.
Van der Waerden theorem (1927): For all k and r there exists n such that in any r-coloring
of [n] there is a monochromatic arithmetic progression of size k.
Erdős-Turán conjecture (1936), Szemerédi theorem (1975) (Roth proved it for k = 3 in
1953 and Szemerédi for k = 4 in 1969): For all k and r there exists n such that all H ⊆ [n] with
|H| ≥ n/r contains an arithmetic progression of size k.
Hales Jewett theorem (1963): For all k and r there exists n such that in any r-coloring of [k]n
there is a monochromatic combinatorial line.
Density Hales Jewett theorem by Fürstenberg and Katznelson (1991): For all k and r
there exists n such that all H ⊆ [k]n with |H| ≥ k n /r contains a combinatorial line.
Let us denote the smallest n satisfying the requirements of above theorems V dW (k, r), Sz(k, r),
HJ(k, r) and DHJ(k, r), respectively. We can consider the theorems to state that these numbers
are finite.
— The density version implies the coloring version, namely V dW ((k, r) ≤ Sz(k, r) and HJ(k, r) ≤
DHJ(k, r). Indeed, the largest color class must contain at least a 1/r fraction of all elements,
look for a monochromatic structure in that color.
— The combinatorial line version implies the arithmetic progression version, namely V dW (k, r) ≤
k HJ(k,r) and Sz(k, r) ≤ k DHJ(k,r) . To see this let us redefine [n] temporarily to be the numbers
from 0 to n − 1. This does not change the problems we consider. Now we can make a 1-1
correspondence between [k m ] and [k]m by writing the numbers in [k m ] as m-digit numbers in
base k (possibly with leading zeros). The inequalities follow since all the combinatorial lines in
[k]m correspond to arithmetic progressions. Note, however, that many arithmetic progressions
do not correspond to combinatorial lines.
The proofs of the VdW and HJ theorems are tricky inductions, but are relatively simple. The
proofs of their density versions are much more involved. By now there are several proofs for both
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density theorems above, one of the simplest ones is a proof of DHJ that was a result of many
mathematicians on-line collaboration in a so called Polymath project in 2009.
Here we give a proof of the Hales Jewett theorem using (nested layers of) induction.
We start with extending the notion of a combinatorial line. For d ≤ n a d-space in [k]n is
an injective map f : [k]d → [k]n such that for all j ∈ [n] its projection fj : [k]d → [k] given by
fj (x) = (f (x))j is either a constant function or one of the coordinate functions x 7→ xi . Note that
a combinatorial line is the image of a 1-space. Here is an example of a 3-space in [3]8 :
f (x, y, z) = (1, z, 3, x, x, z, 3, y).
We call a d-space monochromatic if its image is monochromatic. In particular, (the image of)
a monochromatic 1-space is a monochromatic combinatorial line.
An important but simple feature of d-spaces is that they can be composed: if f : [k]d → [k]m
is a d-space and g : [k]m → [k]n is an m-space, then g ◦ f : [k]d → [k]n is a d-space.
For the inductive proof to work we generalize the Hales-Jewett theorem from monochromatic
combinatorial lines to monochromatic d-spaces. The d = 1 special case of this generalized version
gives the original Hales-Jewett theorem.
Generalized Hales-Jewett theorem: For all k, d and r there exists n such that in any r-coloring
of [k]n there is a monochromatic d-space.
Let GHJ(k, d, r) stand for the smallest n that can be chosen in the statement of this theorem
for k, d and r. We want to prove that this is finite for all k, d and r. For this we prove the following
two lemmas.
z
Lemma 1. For all k, r and d, if z = GHJ(k, d, r) and w = GHJ(k, 1, rk ) are finite, then we have
GHJ(k, d + 1, r) ≤ z + w.
z
Proof: We set r∗ = rk , n = z + w. For x ∈ [k]z and y ∈ [k]w we denote their concatenation
by xy. This is the element of [k]n whose first z coordinates gives x and last w coordinates give y.
Similarly, for x ∈ [k]d and i ∈ [k] we write xi for their concatenation, an element of [k]d+1 .
Let c : [k]n → [r] be an arbitrary r-coloring of [k]n . For y ∈ [k]w we let c∗ (y) : [k]z → [r] be
the r-coloring of [k]z defined by c∗ (y)(x) = c(xy). (This is the restriction of c to the points with
the last w coordinates fixed.) Note that there are r∗ possible r-colorings of [k]z , thus c∗ can be
considered an r∗ -coloring of [k]w . (In a true r∗ -coloring the colors are the elements of [r∗ ] but the
actual values of the colors do not matter, just the number of colors used.)
By the choice of w there is a monochromatic 1-space f : [k] → [k]w for the coloring c∗ . Let
us denote its color by c0 , so c∗ (f (i)) = c0 for all i ∈ [k]. Recall that c0 is actually a function,
c0 : [k]z → [r]. By the choice of z there is a monochromatic d-space g : [k]d → [k]z in [k]z
for the r-coloring c0 . Let t be the color of the image of g. We define h : [k]d+1 → [k]n by
h(x1 , . . . , xd+1 ) = g(x1 , . . . , xd )f (xd+1 ). This is clearly a (d + 1)-space in [k]n . We claim it is
monochromatic in color t. Indeed, for any x ∈ [k]d and i ∈ [k] we have
c(h(xi)) = c(g(x)f (i)) = c∗ (f (i))(g(x)) = c0 (g(x)) = t.
We found a monochromatic (d + 1)-space in any r-coloring of [k]n proving the lemma. Q.E.D.
Lemma 2. For any k and r ≥ 2 if d = GHJ(k + 1, 1, r − 1) + 1 is finite, then GHJ(k + 1, 1, r) ≤
GHJ(k, d, r).
Proof: Let n = GHJ(k, d, r) and let c : [k + 1]n → [r] be an arbitrary r-coloring. The
restriction c0 of c to [k]n is also an r-coloring, so by the choice of n we have a monochromaic
d-space f 0 : [k]d → [k]n . Let t be the color of the image of f 0 and let f : [k + 1]d → [k + 1]n be
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a d-space whose restriction to [k]d is f 0 . (Note that f is unique for k > 1.) For x ∈ [k]d we have
f (x) = t, but f is not necessarily monochromatic on the rest of [k + 1]d . We distinguish two cases
as follows.
Case 1. If there is a point x ∈ [k + 1]d − [k]d with c(f (x)) = t we find a monochromatic 1-space
containing f (x). Indeed, there is a 1-space g : [k + 1] → [k + 1]d such that g(i) ∈ [k]d for i ∈ [k]
and g(k + 1) = x. Now f ◦ g : [k + 1] → [k + 1]n is a monchromatic 1-space in color t.
Case 2. Assume that for x ∈ [k+1]d −[k]d we have c(f (x)) 6= t. Let us define c0 : [k+1]d−1 → [r]
by c0 (x) = c(f (x, k + 1)), where (x, k + 1) is obtained from x ∈ [k + 1]d−1 by attaching k + 1 as the
last coordinate. By our assumption this r-coloring avoids color t, so it is really an (r − 1)-coloring.
By the choice of d this coloring has a monochromatic 1-space g : [k + 1] → [k + 1]d−1 . Clearly, the
1-space h : [k + 1] → [k + 1]n defined by h(i) = f (g(i), k + 1) is monochromatic for the coloring c
(in the same color as the image of g in c0 ).
We found a monochromatic 1-space in both cases proving the lemma. Q.E.D.
Proof of the generalized Hales Jewett theorem: We prove the finitness of GHJ(k, d, r) by
induction on k.
(1) The base case k = 1 is trivial: everything is monochromatic if you only color a single
element, so GHJ(1, d, r) = d.
(2) For k > 1 we prove the finitness of GHJ(k, d, r) by induction on d.
(2/1) The base case d = 1. We prove the finiteness of GHJ(k, 1, r) by induction on r.
(2/1/1) The base case r = 1 is trivial: everything is monochromatic in a 1-coloring, so
GHJ(k, 1, 1) = 1.
(2/1/2) The general case r > 1 follows from Lemma 2: GHJ(k, 1, r) ≤ GHJ(k − 1, d, r) for
d = GHJ(k, 1, r − 1) + 1.
(2/2) The general case d + 1 > 1 follows from Lemma 1: GHJ(k, d + 1, r) ≤ z + w, where
z
z = GHJ(k, d, r) and w = GHJ(k, 1, rk ).
Q.E.D.
Behrend’s construction for sets without 3-term arithmetic progression.
In this last part of these notes we construct a (relatively) large sets without 3-term arithmetic
progression. Note that by Szemerédi’s theorem these sets must contain o(n) elements from [n].
Behrend’s construction gives close to linear number of elements, namely n1−o(1) if t and m is
chosen carefully..
Let n = tm and (once again) identify [n] with the integers 0 ≤ x < n (instead of 1 ≤ x ≤ n).
We write such a number x as a sequence of m digits xi in base t. For 0 ≤ d < m(t/2)2 we write
Hc for the set
X
Hd = {x | xi < t/2,
x2i = c}
The union of these sets contain at least (t/2)m elements, so one of the sets contain at least
(t/2)m /(m(t/2)2 ) numbers.
We claim that none of the sets Hd contain a three term arithmetic progression a, b, c. Indeed,
if it contained one, we had a + c = 2b. Because all digits are small, there is no “carry” in the
addition, so we must have ai + ci = 2bi for the digits of the numbers. This implies a2i + c2i ≥ 2b2i .
Summing this inequality for all i we get 2d on boh sides, so equality must hold for all i. This
implies ai = bi = ci for all i, so a = b = c contradicting the assumption that the number form an
arithmetic progression.
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