Gallai’s Theorem: A Detailed Exposition of Witt’s Proof

Abstract

Gallai’s theorem states that if the Euclidean plane is colored with finitely many colors then every finite configuration of points admits a monochromatic homothetic copy. This paper presents a detailed exposition of a proof originally published by Ernst Witt in 1952 and subsequently expanded by Alexander Soifer. Additional intermediate steps are supplied throughout, yielding a self-contained and easily verified proof. The argument is formulated recursively through finite configurations associated with a double induction and thereby makes explicit the finite structures underlying the theorem.

1. Introduction

A theorem of Tibor Gallai (formerly Grünwald) states that if the points in the Euclidean plane are colored with finitely many colors, then for every finite subset of the plane there is a monochromatic homothetic copy of that set. A homothetic copy of a set is its image under first a dilation and then a translation. A monochromatic copy is one in which every point receives the same color. Another way to state Gallai’s theorem is that for every finite subset X of the plane and every number k of colors, there is another finite subset Y of the plane, such that if Y is colored by k colors then Y contains a monochromatic homothetic image of X. Gallai’s Theorem can be stated as follows as a higher-dimensional generalization of van der Waerden’s theorem, where $\mathbb{E}$ is any power of $\mathbb{R}$ (or even $\mathbb{Q}$ or $\mathbb{Z}$):

Theorem 1.

For $m,n,k$ arbitrary positive integers, if the Euclidean space $\mathbb{E}$ is colored with k colors and $S_n\subseteq \mathbb{E}$ an n-element subset, there exists a monochromatic subset $S_n^{\prime }\subseteq \mathbb{E}$ which is homothetic to $S_n$.

Gallai did not publish his result. (For an explanation and additional history, one should read both [1] (Chapter 42) and [2] (Chapter 46).) Gallai’s theorem was first mentioned in print by Richard Rado [3]. Reviewing Rado’s paper [3] for Mathematical Reviews, Erdős wrote, “the following result of T. Grünwald is used: given any configuration S consisting of a finite number of lattice points of Euclidean space, and given a distribution of all lattice points of this space into a finite number of classes, there is at least one class which contains a configuration $S^{\prime }$ of lattice points which is similar and parallel to S.”

I first heard about Gallai’s theorem in late 2012 from Jeremy Alm, who wrote and sent me a paper about extending Gallai’s theorem [4]. I became intrigued by the theorem, tried to prove it, and eventually consulted the reference in Jeremy’s paper, Alexander Soifer’s The Mathematical Coloring Book [1] (Section 42.3). Soifer presents an expanded account of the proof of Gallai’s theorem published by Ernst Witt in 1952 [5]. Reviewing Witt’s paper [5] for Mathematical Reviews, Erdős wrote, “The author was unaware of a paper by R. Rado in which it is stated that the result is due to T. Grünwald (Gallai).” Soifer thought Witt’s proof was incomprehensibly brief, so he added details but did not include the proof in the 2024 edition [2]. Soifer’s proof seemed incomprehensibly brief to me, so I added even more details, enough for a purely formal confirmation of the proof.

The theorem and its proof are written for ${\mathbb{R}}^2$, but any power of $\mathbb{R}$ (or $\mathbb{Q}$ or $\mathbb{Z}$) can be substituted for ${\mathbb{R}}^2$. The elements are either called points or vectors, depending on their use as either geometric or computational objects.

The theorem is proved in a slightly more precise formulation than the usual statement. For an infinite set of distinct points $\{e_0,e_1,\cdots \}\subseteq {\mathbb{R}}^2$, an integer $n\ge 1$, and $k\ge 1$ indicating the number of colors, a set $\mathsf{\Phi }(n,k)$ is defined recursively in advance with the property that for any coloring of $\mathsf{\Phi }(n,k)$ with k colors, $\mathsf{\Phi }(n,k)$ contains a monochromatic homothetic copy of $\{e_0,\dots ,e_{n-1}\}$. The key idea, as in many combinatorial arguments, is to treat entire colorings as colors and to apply recursive coloring constructions within the inductive argument.

2. Notations

The notational conventions listed here are followed throughout the paper. $\mathbb{R}$ will be considered as our ground field. The symbols $\lambda$ and $\mu$, with or without subscripts, denote elements of $\mathbb{R}$. We use i, j, k, ℓ, m, and n as numerical parameters in $\mathbb{N}$, subject to these interpretations: k is the number of colors, n is the number of points, and the meanings of the others depend on context. The elements of ${\mathbb{R}}^2$—called points or vectors depending on context—are a, b, v, w, $e_0$, $e_1$, etc. The symbol + denotes addition of numbers, or vectors, or sets of vectors. The symbols that denote homotheties (compositions of translations and dilations mapping ${\mathbb{R}}^2$ to ${\mathbb{R}}^2$) are h, $h^{\prime }$, and g, with or without subscripts. For colorings we use f and $f^{\prime }$. For n points and k colors, the notations used for subsets of ${\mathbb{R}}^2$ are V, $E_n\left(V\right)$, S, $S^{\prime }$, $S_n$, $S_n^{\prime }$, $\mathsf{\Phi }(n,k)$, and $\Delta (n,k,m)$. $A(n,k)$ and $D(n,k,m)$ are certain statements about $\mathsf{\Phi }(n,k)$ and $\Delta (n,k,m)$.

3. Basic Structures

Consider an arbitrary but fixed infinite sequence $S=〈e_0,e_1,e_2,\cdots ,e_i,\cdots 〉$ of distinct points in ${\mathbb{R}}^2$, where $e_0=(0,0)$. For every integer $n\ge 1$ let

$$S_n=\{e_0,\cdots ,e_{n-1}\}.$$

(1)

For example, $S_1=\{e_0\}$, $S_2=\{e_0,e_1\}$, and $S_3=\{e_0,e_1,e_2\}$. For every set X, $\left|X\right|$ is the number of elements of X, called the cardinality of X. The cardinality of $S_n$ is n, i.e., $|S_n|=n$. For every positive integer k, a k-element set is a set X such that $\left|X\right|=k$. A function from a set Y to a k-element set X is a k-coloring of Y. For any vector $v\in {\mathbb{R}}^2$ and any two sets $V,W\subseteq {\mathbb{R}}^2$, the sumset of v (or V) and W is the set $v+W$ (or $V+W$) of vectors obtained by adding v (or any vector from V) to any vector from W, that is,

$$\begin{array}{cc}\hfill v+W& =\{v+w:w\in W\},\hfill \\ \hfill V+W& =\{v+w:v\in V,w\in W\}.\hfill \end{array}$$

Let H be the set of all functions $h:{\mathbb{R}}^2\to {\mathbb{R}}^2$ such that, for some displacement vector $a\in {\mathbb{R}}^2$ and some positive dilation scalar $\lambda \in \mathbb{R}$, we have $h\left(v\right)=a+\lambda v$ for every $v\in {\mathbb{R}}^2$, where addition and scalar multiplication are performed componentwise. H is the set of homotheties from ${\mathbb{R}}^2$ to ${\mathbb{R}}^2$, consisting of all compositions of translations and dilations, i.e.,

$$H=\{h:{\mathbb{R}}^2\to {\mathbb{R}}^2|\exists a\in {\mathbb{R}}^2\mathrm{and}\exists \lambda \in \mathbb{R}\mathrm{such}\mathrm{that}h\left(v\right)=a+\lambda v\mathrm{for}\mathrm{all}v\in {\mathbb{R}}^2\}.$$

If $V\subseteq {\mathbb{R}}^2$ and $h\in H$, then $h\left(V\right)$ is the image of V under h, i.e.,

$$h\left(V\right)=\{h\left(v\right):v\in {\mathbb{R}}^2\}.$$

For $V\subseteq {\mathbb{R}}^2$ arbitrary and $n\ge 2$ integer, the set $E_n\left(V\right)$ will be defined as the union of the images of $S_n$ under homotheties $h\in H$ mapping $S_{n-1}$ into V, that is

$$E_n\left(V\right)=\bigcup _n\left\{h\left(S_n\right):h\in H\mathrm{and}h\left(S_{n-1}\right)\subseteq V\right\}.$$

(2)

For $n\ge 1$, $k\ge 1$, and $m\ge 1$ integers, define subsets

$$\mathsf{\Phi }(n,k),\Delta (n,k,m)\subseteq {\mathbb{R}}^2$$

by using the following recursive relations:

$$\begin{array}{cc}\hfill \mathsf{\Phi }(1,k)& =\{e_0\},\hfill \end{array}$$

(3)

$$\begin{array}{cc}\hfill \mathsf{\Phi }(2,k)& =\{e_0,e_1,2e_1,\dots ,k{e}_1\},\hfill \end{array}$$

(4)

$$\begin{array}{cc}\hfill \Delta (n,k,1)& =E_n\left(\mathsf{\Phi }(n-1,k)\right),\hfill \end{array}$$

(5)

$$\begin{array}{cc}\hfill \Delta (n,k,m+1)& =\Delta (n,k^{|\Delta (n,k,m\left)\right|},1)+\Delta (n,k,m),\hfill \end{array}$$

(6)

$$\begin{array}{cc}\hfill \mathsf{\Phi }(n,k)& =\Delta (n,k,k).\hfill \end{array}$$

(7)

For example, with $n=3$, $k=2$, and $m=1$, we have $\mathsf{\Phi }(2,2)=\{e_0,e_1,2e_1\}$ by (4), so by (5),

$$\begin{array}{c}\hfill \Delta (3,2,1)=E_3\left(\mathsf{\Phi }(2,2)\right)=E_3\left(\{e_0,e_1,2e_1\}\right).\end{array}$$

There are three homotheties $h_1,h_2,h_3\in H$ that map $S_2=\{e_0,e_1\}$ into $\{e_0,e_1,2e_1\}$, defined for all $v\in {\mathbb{R}}^2$ by $h_1\left(v\right)=v$, $h_2\left(v\right)=2v,$ and $h_3\left(v\right)=e_1+v.$ To check this, recall that $e_0=(0,0)$. Then

$$\begin{array}{ccc}\hfill E_3\left(\{e_0,e_1,2e_1\}\right)& =\bigcup \{h\left(S_3\right):h\in H,h\left(S_2\right)\subseteq \{e_0,e_1,2e_1\}\}\hfill & \mathrm{by}\left(2\right)\hfill \\ & =\bigcup \{h\left(S_3\right):h\in \{h_1,h_2,h_3\}\}\hfill & \\ & =h_1\left(S_3\right)\cup h_2\left(S_3\right)\cup h_3\left(S_3\right)\hfill & \\ & =h_1\left(\{e_0,e_1,e_2\}\right)\cup h_2\left(\{e_0,e_1,e_2\}\right)\cup h_3\left(\{e_0,e_1,e_2\}\right)\hfill & \mathrm{by}\left(1\right)\hfill \\ & =\{e_0,e_1,e_2\}\cup \{e_0,2e_1,2e_2\}\cup \{e_1,2e_1,e_1+e_2\}\hfill & \\ & =\{e_0,e_1,e_2,2e_1,2e_2,e_1+e_2\},\hfill & \end{array}$$

so $|\Delta (3,2,1\left)\right|\le 6$, and for the case when $|\Delta (3,2,1\left)\right|=6$, we have

$$\begin{array}{ccc}\hfill \mathsf{\Phi }(3,2)& =\Delta (3,2,2)\hfill & \mathrm{by}\left(7\right)\hfill \\ & =\Delta (3,2^{|\Delta (3,2,1\left)\right|},1)+\Delta (3,2,1)\hfill & \mathrm{by}\left(6\right)\hfill \\ & =\Delta (3,2^6,1)+\{e_0,e_1,e_2,2e_2,e_1+e_2\}\hfill & \\ & =E_3\left(\mathsf{\Phi }(2,64)\right)+\{e_0,e_1,e_2,2e_2,e_1+e_2\}\hfill & \mathrm{by}(5).\hfill \end{array}$$

Since $\mathsf{\Phi }(2,64)$ has 65 points by (4), there are $\left({\displaystyle \genfrac{}{}{0pt}{}{65}{2}}\right)=2080$ homotheties mapping $S_2$ into $\mathsf{\Phi }(2,64)$; hence $E_3\left(\mathsf{\Phi }(2,64)\right)$ may contain as many as $65+2080=2145$ points, and $\mathsf{\Phi }(3,2)$ may contain as many as $2145·6=\mathrm{12,870}$.

4. Gallai’s Theorem

Following the multi-dimensional version of Gallai’s Theorem, mentioned in the introduction (Theorem 1), the two dimensional case could be phrased as follows, which is quoted from [1].

Theorem: For any arbitrary positive integer k and a finite n-element set $S_n=\{e_0,\dots ,e_{n-1}\}$ of points in the plane, there is a finite set $\mathsf{\Phi }=\mathsf{\Phi }(n,k)$ of points in the plane such that if $\mathsf{\Phi }$ is colored in k colors, there is a monochromatic subset $S_n^{\prime }$ of $\mathsf{\Phi }$ which is homothetic to $S_n$.

This result has already been proven by Witt [5] and later a detailed version of Witt’s proof appeared in Soifer’s book [1]. The main aim is to provide an even more comprehensively detailed proof based on Soifer’s proof. To do this, as in Soifer’s version, we will prove both of the following statements at the same time.

$A(n,k)$: For every k-coloring of $\mathsf{\Phi }(n,k)$, $\mathsf{\Phi }(n,k)$ contains a monochromatic homothetic image of $S_n$.

$D(n,k,m)$: For every k-coloring f of $\Delta (n,k,m)$, there are scalars ${\lambda }_0,\dots ,{\lambda }_m\in \mathbb{R}$ and a vector $a\in {\mathbb{R}}^2$ such that $0={\lambda }_0<{\lambda }_1<\cdots <{\lambda }_m$, and if $h_{i,j}\left(v\right)=a+{\lambda }_i{e}_n+({\lambda }_j-{\lambda }_i)v$ whenever $v\in {\mathbb{R}}^2$ and $0\le i<j\le m$, then $h_{i,j}\left(S_n\right)\subseteq \Delta (n,k,m)$ and $\left|f\right(h_{i,j}\left(S_{n-1}\right)\left)\right|=1$.

As in [1], instead of Gallai’s Theorem, we will prove the following equivalent result.

Theorem 2.

For $n\ge 1$, $k\ge 1$, and $m\ge 1$ integers, we have $A(n,k)$ and $D(n,k,m)$.

The proof will proceed by a double induction on m and n. Consider Equations (4)–(7), and suppose that the right-hand sides of these equations possess the requisite properties of statements $A(n,k)$ and $D(n,k,m)$. Then we will show that the left-hand sides do so as well. In other words, for $n,k,m$ as before, the plan is to show that the following statements hold.

Induction steps:

$$A(1,k),$$

(8)

$$\begin{array}{c}\hfill \mathrm{if}n\ge 2\mathrm{and}A(n-1,k) \mathrm{then} D(n,k,1),\end{array}$$

(9)

$$\mathrm{if}D\left(n,k,m\right) \mathrm{and} D(n,k^{|\Delta (n,k,m\left)\right|},1), \mathrm{then} D(n,k,m+1),$$

(10)

$$\mathrm{if}D\left(n,k,k\right) \mathrm{then} A(n,k).$$

(11)

To see that these statements are enough to establish the theorem, we explicitly lay out the initial inductive steps. By (8), we have

(i)

$A(2,k)$ for every $k\ge 1$.

For every $k\ge 2$ we get $D(3,k,1)$ by (i) and (9) with $n=3$. This proves

(ii)

$D(3,k,1)$ for every $k\ge 1$.

For every $k\ge 2$, we get $D(3,k^{|\Delta (3,k,1\left)\right|},1)$ and $D(3,k,1)$ from (ii), hence $D(3,k,2)$ by (10) with $n=3$ and $m=1$. This proves

(iii)

$D(3,k,2)$ for every $k\ge 1$.

For every $k\ge 2$, we get $D(3,k^{|\Delta (3,k,2\left)\right|},1)$ by (ii) and $D(3,k,2)$ by (iii), and hence $D(3,k,3)$ by (10) with $n=3$ and $m=2$. This proves

(iv)

$D(3,k,3)$ for every $k\ge 1$.

Taking (ii) and (iii) as base cases and the reasoning from (iii) to (iv) as the inductive step, we obtain a proof by induction that

(v)

$D(3,k,m)$ for every $k\ge 1$ and $m\ge 1$.

For every $k\ge 1$, we have $D(3,k,k)$ by (v). It follows by (11) with $n=3$ that

(vi)

$A(3,k)$ for every $k\ge 1$.

The deduction from (i) to (vi) is just the first step (from $n=2$ to $n=3$) in an inductive proof that

(vii)

$A(n,k)$ for every $k\ge 2$ and every $n\ge 2$.

Note that (vii) is one way to state Gallai’s theorem, but (i)–(vii) incorporate all the statements about $\Delta$ mentioned in the exposition of the double induction.

5. Proof of $\mathit{A}(\mathbf{1},\mathit{k})$ and $\mathit{A}(\mathbf{2},\mathit{k})$ for $\mathit{k}\ge \mathbf{1}$

Assume $k\ge 1$. $A(1,k)$ holds in a very strong way, for if the space is colored with any number of colors, even infinitely many, then $\mathsf{\Phi }(1,k)=\{e_0\}$ contains monochromatic homothetic images of $S_1=\{e_0\}$. We turn to $A(2,k)$. Since $e_0=(0,0)$, definition (4) implies that

$$\begin{array}{c}\hfill \mathsf{\Phi }(2,k)=\{e_0,e_1,2e_1,\cdots ,k{e}_1\}.\end{array}$$

Assume that f is a k-coloring of $\mathsf{\Phi }(2,k)$. We must show that $\mathsf{\Phi }(2,k)$ contains a monochromatic homothetic image of $S_2=\{e_0,e_1\}$. From $e_0\ne e_1$, it follows that the number of points in $\mathsf{\Phi }(2,k)$ is $k+1$, one more than the number of colors, so there are (at least) two points in $\mathsf{\Phi }(2,k)$ that obtain the same color, say $f\left(i{e}_1\right)=f\left(j{e}_1\right)$ for some $i,j$, such that $0\le i<j\le k$. Define h by $h\left(v\right)=i{e}_1+(j-i)v$ for all $v\in {\mathbb{R}}^2$. Then, $h\in H$ since $j-i>0$ and h maps $S_2=\{e_0,e_1\}$ to $\{i{e}_1,j{e}_1\}\subseteq \mathsf{\Phi }(2,k)$ because

$$\begin{array}{cc}\hfill h\left(e_0\right)& =i{e}_1+(j-i)e_0=i{e}_1+(j-i)(0,0)=i{e}_1,\hfill \\ \hfill h\left(e_1\right)& =i{e}_1+(j-i)e_1=i{e}_1+j{e}_1-i{e}_1=j{e}_1.\hfill \end{array}$$

This homothetic image is monochromatic because $f\left(i{e}_1\right)=f\left(j{e}_1\right)$.

6. Proof That If $\mathit{n}\ge \mathbf{3}$, $\mathit{k}\ge \mathbf{1}$, and $\mathit{A}(\mathit{n}-\mathbf{1},\mathit{k})$, Then $\mathit{D}(\mathit{n},\mathit{k},\mathbf{1})$

Assume $n\ge 3$, $k\ge 1$, and $A(n-1,k)$. Assume f is a k-coloring of $\Delta (n,k,1)$. By the assumption $A(n-1,k)$, $\mathsf{\Phi }(n-1,k)$ contains a monochromatic homothetic image of $S_{n-1}$, so there is some $h\in H$ such that

$$\begin{array}{cc}\hfill & h\left(S_{n-1}\right)\subseteq \mathsf{\Phi }(n-1,k),\hfill \end{array}$$

(12)

$$\begin{array}{cc}\hfill & \left|f\right(h\left(S_{n-1}\right)\left)\right|=1.\hfill \end{array}$$

(13)

From (12), the definition of $E_n$, and definition (5), it follows that

$$h\left(S_n\right)\subseteq E_n\left(\mathsf{\Phi }(n-1,k)\right)=\Delta (n,k,1).$$

(14)

To conclude that $D(n,k,1)$, we must find ${\lambda }_0,{\lambda }_1\in \mathbb{R}$ and $a\in {\mathbb{R}}^2$ such that $0={\lambda }_0<{\lambda }_1$, $h_{0,1}\left(S_n\right)\subseteq \Delta (n,k,1)$, and $\left|f\right(h_{0,1}\left(S_{n-1}\right)\left)\right|=1$, where

$$h_{0,1}\left(v\right)=a+{\lambda }_0{e}_n+({\lambda }_1-{\lambda }_0)v=a+{\lambda }_{1}v$$

whenever $v\in {\mathbb{R}}^2$. It suffices to let ${\lambda }_1$ and a be the dilation scalar and displacement vector associated with h, for then $h_{0,1}=h$ and the two desired equations are (13) and (14).

7. Proof of $\mathit{D}(\mathit{n},\mathit{k},\mathit{m}+\mathbf{1})$ from $\mathit{D}(\mathit{n},\mathit{k},\mathit{m})$ and $\mathit{D}(\mathit{n},{\mathit{k}}^{|\Delta (\mathit{n},\mathit{k},\mathit{m}\left)\right|},\mathbf{1})$

Suppose f is a k-coloring of $\Delta (n,k,m+1)$. By definition (6), f assigns a color to every vector obtained by adding a vector from $\Delta (n,k^{|\Delta (n,k,m\left)\right|},1)$ to a vector in $\Delta (n,k,m)$. Therefore, for every $v\in \Delta (n,k^{|\Delta (n,k,m\left)\right|},1)$, we may let $f^{\prime }\left(v\right)$ be the k-coloring of $\Delta (n,k,m)$ defined by

$$f^{\prime }\left(v\right)\left(w\right)=f(v+w)\mathrm{for}\mathrm{every}w\in \Delta (n,k,m).$$

(15)

This gives us a new coloring $f^{\prime }$ that assigns each vector $v\in \Delta (n,k^{|\Delta (n,k,m\left)\right|},1)$ to an element $f^{\prime }\left(v\right)$ of the set of k-colorings of $|\Delta (n,k,m\left)\right|$, a set whose cardinality is $k^{|\Delta (n,k,m\left)\right|}$. Thus $f^{\prime }$ is a $k^{|\Delta (n,k,m\left)\right|}$-coloring of $\Delta (n,k^{|\Delta (n,k,m\left)\right|},1)$. From the inductive hypothesis $D(n,k^{|\Delta (n,k,m\left)\right|},1)$, applied to the coloring $f^{\prime }$, we know there are ${\mu }_0,{\mu }_1\in \mathbb{R}$ and $b\in {\mathbb{R}}^2$ such that $0={\mu }_0<{\mu }_1$ and, defining $g_{0,1}$ by

$$g_{0,1}\left(v\right)=b+{\mu }_0{e}_n+({\mu }_1-{\mu }_0)v=b+{\mu }_{1}v\mathrm{for}\mathrm{all}v\in {\mathbb{R}}^2,$$

(16)

we have

$$\begin{array}{cc}\hfill & g_{0,1}\left(S_n\right)\subseteq \Delta (n,k^{|\Delta (n,k,m\left)\right|},1),\hfill \end{array}$$

(17)

$$\begin{array}{cc}\hfill & |f^{\prime }\left(g_{0,1}\left(S_{n-1}\right)\right)|=1.\hfill \end{array}$$

(18)

From the inductive hypothesis $D(n,k,m)$, applied to the k-coloring $f^{\prime }\left(b\right)$, we know there are ${\lambda }_0,{\lambda }_1,\cdots ,{\lambda }_m\in \mathbb{R}$ and $a\in {\mathbb{R}}^2$ such that

$$0={\lambda }_0<{\lambda }_1<\cdots <{\lambda }_m$$

(19)

and, assuming

$$h_{i,j}\left(v\right)=a+{\lambda }_i{e}_n+({\lambda }_j-{\lambda }_i)v\mathrm{whenever}v\in {\mathbb{R}}^2\mathrm{and}0\le i<j\le m,$$

(20)

we have

$$\begin{array}{cc}\hfill & h_{i,j}\left(S_n\right)\subseteq \Delta (n,k,m),\hfill \end{array}$$

(21)

$$\begin{array}{cc}\hfill & |f^{\prime }\left(b\right)\left(h_{i,j}\left(S_{n-1}\right)\right)|=1.\hfill \end{array}$$

(22)

Our next plan is to prove, assuming $w\in \Delta (n,k,m)$, $0\le \mathsf{\ell }<n-1$, and $0\le i<m$, that the equations

$$\begin{array}{cc}& f(b+w)=f(b+{\mu }_1{e}_{\mathsf{\ell }}+w),\hfill \end{array}$$

(23)

$$\begin{array}{cc}& f(b+a+{\lambda }_i{e}_n)=f(b+a+{\lambda }_i{e}_n+({\lambda }_m-{\lambda }_i)e_{\mathsf{\ell }}),\hfill \end{array}$$

(24)

hold. Note that $e_0,e_{\mathsf{\ell }}\in S_{n-1}$ beause $0\le \mathsf{\ell }<n-1$. By using (16), it is not too difficult to see that $g_{0,1}\left(e_0\right)=b$ and $g_{0,1}\left(e_{\mathsf{\ell }}\right)=b+{\mu }_1{e}_{\mathsf{\ell }}$. Therefore, $b,b+{\mu }_1{e}_{\mathsf{\ell }}\in g_{0,1}\left(S_{n-1}\right)$. Using (18) we can conclude that $f^{\prime }\left(b\right)=f^{\prime }(b+{\mu }_1{e}_{\mathsf{\ell }})$, and by using (15) and $w\in \Delta (n,k,m)$ we complete the proof of (23).

To prove (24), we will follow almost the same approach as that which we used to prove (23). We have $e_0,e_{\mathsf{\ell }}\in S_{n-1}$ because $0\le \mathsf{\ell }<n-1$ and (20) can be applied with $j=m$ because $0\le i<m$, hence

$$\begin{array}{cc}\hfill & a+{\lambda }_i{e}_n=h_{i,m}\left(e_0\right)\in h_{i,m}\left(S_{n-1}\right),\hfill \\ & a+{\lambda }_i{e}_n+({\lambda }_m-{\lambda }_i)e_{\mathsf{\ell }}=h_{i,m}\left(e_{\mathsf{\ell }}\right)\in h_{i,m}\left(S_{n-1}\right).\hfill \end{array}$$

By using (21), we see that both of these points are in $\Delta (n,k,m)$ and get the same color from $f^{\prime }\left(b\right)$ by (22) with $j=m$, that is,

$$f^{\prime }\left(b\right)(a+{\lambda }_i{e}_n)=f^{\prime }\left(b\right)(a+{\lambda }_i{e}_n+({\lambda }_m-{\lambda }_i)e_{\mathsf{\ell }}).$$

To conclude the proof of (24), apply (15) with $v=b$.

To continue, we are required to define a new system as follows. Let ${\lambda }_{m+1}={\mu }_1+{\lambda }_m$. For $v\in {\mathbb{R}}^2$ and $0\le i<j\le m+1$ define

$$h_{i,j}^{\prime }\left(v\right)=b+a+{\lambda }_i{e}_n+({\lambda }_j-{\lambda }_i)v.$$

(25)

For ${\mu }_1$ as before, define ${\lambda }_{m+1}={\lambda }_m+{\mu }_1$. As ${\mu }_1>0$, hence ${\lambda }_{m+1}>{\lambda }_m$ is apparent, and so the inequality

$$0={\lambda }_0<{\lambda }_1<\cdots <{\lambda }_m<{\lambda }_{m+1}$$

holds. Now, for $v\in {\mathbb{R}}^2$ and $0\le i<j\le m+1$, in order to conclude $D(n,k,m+1)$ for the newly defined system $h_{i,j}^{\prime }$, we need to show that the following equations hold:

$$\begin{array}{cc}\hfill & h_{i,j}^{\prime }\left(S_n\right)\subseteq \Delta (n,k,m+1),\hfill \end{array}$$

(26)

$$\begin{array}{cc}\hfill & \left|f\right(h_{i,j}^{\prime }\left(S_{n-1}\right)\left)\right|=1.\hfill \end{array}$$

(27)

Case 1. $0\le i<j\le m$.

$$\begin{array}{cccc}\hfill h_{i,j}^{\prime }\left(S_n\right)& =b+h_{i,j}\left(S_n\right)\hfill & \hfill & \mathrm{by}\left(20),\right(25)\hfill \\ & \subseteq b+\Delta (n,k,m)\hfill & \hfill & \mathrm{by}\left(21\right)\hfill \\ & =g_{0,1}\left(e_0\right)+\Delta (n,k,m)\hfill & \hfill & \mathrm{by}(16),e_0=(0,0)\hfill \\ & \subseteq g_{0,1}\left(S_n\right)+\Delta (n,k,m)\hfill & \hfill & e_0\in S_n\hfill \\ & \subseteq \Delta (n,k^{|\Delta (n,k,m\left)\right|},1)+\Delta (n,k,m)\hfill & \hfill & \mathrm{by}\left(17\right)\hfill \\ & =\Delta (n,k,m+1)\hfill & \hfill & \mathrm{by}(6).\hfill \end{array}$$

So (26) holds in this case, and as we have

$$\begin{array}{cccc}\hfill \left|f\right(h_{i,j}^{\prime }\left(S_{n-1}\right)\left)\right|& =\left|f\right(b+h_{i,j}\left(S_{n-1}\right)\left)\right|\hfill & \hfill & \mathrm{by}\left(20),\right(25)\hfill \\ & =|f^{\prime }\left(b\right)\left(h_{i,j}\left(S_{n-1}\right)\right)|\hfill & \hfill & \mathrm{by}\left(15\right)\hfill \\ & =1\hfill & \hfill & \mathrm{by}(22),\hfill \end{array}$$

hence, (27) holds in this case as well.

• Case 2. $0\le i<j=m+1$.

If $v\in S_n$, then

$$\begin{array}{ccc}\hfill h_{i,m+1}^{\prime }\left(v\right)& =b+a+{\lambda }_i{e}_n+({\lambda }_{m+1}-{\lambda }_i)v\hfill & \mathrm{by} \left(25\right)\hfill \\ & =b+a+{\lambda }_i{e}_n+({\lambda }_m+{\mu }_1-{\lambda }_i)v\hfill & \mathrm{by} \mathrm{def}. \mathrm{of} {\lambda }_{m+1}\hfill \\ & =b+{\mu }_{1}v+a+{\lambda }_i{e}_n+({\lambda }_m-{\lambda }_i)v\hfill & \\ & =g_{0,1}\left(v\right)+h_{i,m}\left(v\right)\hfill & \mathrm{by} \left(16\right), \left(20\right)\hfill \\ & \in \Delta (n,k^{|\Delta (n,k,m\left)\right|},1)+\Delta (n,k,m)\hfill & \mathrm{by} \left(17\right), \left(21\right)\hfill \\ & =\Delta (n,k,m+1)\hfill & \mathrm{by} \left(6\right),\hfill \end{array}$$

and this concludes the proof of (26).

Consider $e_{\mathsf{\ell }}\in S_{n-1}$ for arbitrary $0\le \mathsf{\ell }<n-1$. For $j=m$, we have

$$a+{\lambda }_i{e}_n+({\lambda }_m-{\lambda }_i)e_{\mathsf{\ell }}=h_{i,m}\left(e_{\mathsf{\ell }}\right)\in \Delta (n,k,m),$$

(28)

by using (20) and (21), and as $i<m+1$,

$$\begin{array}{ccc}\hfill f\left(h_{i,m+1}^{\prime }\left(e_0\right)\right)& =f(b+a+{\lambda }_i{e}_n)\hfill & \mathrm{by}(25),e_0=(0,0)\hfill \\ & =f(b+a+{\lambda }_i{e}_n+({\lambda }_m-{\lambda }_i)e_l)\hfill & \mathrm{by}\left(24\right)\mathrm{if}i<m,\mathrm{trivial}\mathrm{if}i=m\hfill \\ & =f(b+{\mu }_1{e}_l+a+{\lambda }_i{e}_n+({\lambda }_m-{\lambda }_i)e_l)\hfill & \mathrm{by}\left(28),\right(23)\hfill \\ & =f(b+a+{\lambda }_i{e}_n+({\mu }_1+{\lambda }_m-{\lambda }_i)e_l)\hfill & \\ & =f(b+a+{\lambda }_i{e}_n+({\lambda }_{m+1}-{\lambda }_i)e_l)\hfill & \mathrm{by}\mathrm{definition}\mathrm{of}{\lambda }_{m+1}\hfill \\ & =f\left(h_{i,m+1}^{\prime }\left(e_l\right)\right)\hfill & \mathrm{by}(25),\hfill \end{array}$$

and this concludes the proof of (27) in this case.

8. Proof of $\mathit{A}(\mathit{n},\mathit{k})$ from $\mathit{D}(\mathit{n},\mathit{k},\mathit{k})$

As we have $\mathsf{\Phi }(n,k)=\Delta (n,k,k)$ by definition (7), what we need to show is that $\Delta (n,k,k)$ contains a monochromatic homothetic image of $S_n$, assuming f is a k-coloring of $\Delta (n,k,k)$. From the inductive hypothesis $D(n,k,k)$, we know there are ${\lambda }_0,{\lambda }_1,\cdots ,{\lambda }_k\in \mathbb{R}$ and $a\in {\mathbb{R}}^2$ such that $0={\lambda }_0<{\lambda }_1<\cdots <{\lambda }_k$ and if

$$h_{i,j}\left(v\right)=a+{\lambda }_i{e}_n+({\lambda }_j-{\lambda }_i)v$$

whenever $v\in {\mathbb{R}}^2$ and $0\le i<j\le k$, then $h_{i,j}\left(S_n\right)\subseteq \Delta (n,k,k)$ and $\left|f\right(h_{i,j}\left(S_{n-1}\right)\left)\right|=1$. Let us look at the images of $e_n$ and $e_0$ under all these homotheties $h_{i,j}$ with $0\le i<j\le k$. We have

$$h_{i,j}\left(e_n\right)=a+{\lambda }_i{e}_n+({\lambda }_j-{\lambda }_i)e_n=a+{\lambda }_j{e}_n,0<j\le k,$$

and, since $e_0=(0,0)$,

$$h_{i,j}\left(e_0\right)=a+{\lambda }_i{e}_n+({\lambda }_j-{\lambda }_i)e_0=a+{\lambda }_i{e}_n,0\le i<k.$$

Consequently, the two k-element lists of images of $e_n$ and $e_0$ are, respectively,

$$\begin{array}{cc}\hfill & a+{\lambda }_1{e}_n,\cdots ,a+{\lambda }_{k-1}{e}_n,a+{\lambda }_k{e}_n,\hfill \\ & a,a+{\lambda }_1{e}_n,\cdots ,a+{\lambda }_{k-1}{e}_n.\hfill \end{array}$$

Both lists contain k elements, their intersection has only $k-1$ elements, and their union is a $(k+1)$-element set, to which are assigned only k colors. Therefore, two elements receive the same color: there are $i,j$ such that $0\le i<j\le k$ and

$$f(a+{\lambda }_i{e}_n)=f(a+{\lambda }_j{e}_n).$$

However, $a+{\lambda }_i{e}_n=h_{i,j}\left(e_0\right)$ and $a+{\lambda }_j{e}_n=h_{i,j}\left(e_n\right)$, so

$$f\left(h_{i,j}\left(e_0\right)\right)=f\left(h_{i,j}\left(e_n\right)\right).$$

Of course, we already know $\left|f\right(h_{i,j}\left(S_{n-1}\right)\left)\right|=1$, so this last equation tells us, since $e_0\in S_{n-1}$, that $e_n$ also receives the same color as all the other elements of $S_{n-1}$. But $S_n=S_{n-1}\cup \{e_n\}$, so in fact we have shown $\left|f\right(h_{i,j}\left(S_n\right)\left)\right|=1$. We also know that $h_{i,j}\left(S_n\right)\subseteq \Delta (n,k,k)$, so we conclude that $\Delta (n,k,k)$ does indeed contain a monochromatic homothetic image of $S_n$.

9. Example

The example mentioned immediately after (7) is continued here with an explicit choice for the three points. This will illustrate the enormous numbers of points involved in Witt’s proof. Let $e_0=(0,0)$, $e_1=(1,0)$, and $e_2=(1,1)$. Then $S_3$ forms a right triangle with its right angle at $e_1$. For $1\le k$, (4) implies that

$$\begin{array}{cc}\hfill \mathsf{\Phi }(2,k)& =\{(0,0),(1,0),(2,0),\cdots ,(k,0)\}.\hfill \end{array}$$

There are $k+1$ points in this set. We next compute the set $\Delta (3,k,1)$. By (2) and (5), we have

$$\begin{array}{cc}\hfill \Delta \left(3,k,1\right)& =E_3\left(\mathsf{\Phi }\left(2,k\right)\right)=\bigcup \{h\left(S_3\right):h\in H\mathrm{and}h\left(S_2\right)\subseteq \mathsf{\Phi }\left(2,k\right)\},\hfill \end{array}$$

so we need all the homotheties that map $S_2$ into $\mathsf{\Phi }(2,k)$. Assume $h\left(v\right)=a+\lambda v$, $a\in {\mathbb{R}}^2$, $0<\lambda \in \mathbb{R}$, and h maps $S_2$ into $\mathsf{\Phi }(2,k)$. Then $h\left(e_0\right)=a=(m,0)$ for some $m\in \{0,\dots ,k\}$, hence $h\left(e_1\right)=(m+\lambda ,0)$, but $0<\lambda$ so $m<m+\lambda$, i.e., $h\left(e_1\right)$ is a later point than a in the sequence $(0,0)$, $(1,0)$, $(2,0)$, …, $(k,0)$. Therefore, there are $\left({\displaystyle \genfrac{}{}{0pt}{}{k+1}{2}}\right)$ homotheties that map $S_2$ into $\mathsf{\Phi }(2,k)$. If $h\left(e_1\right)=(\mathsf{\ell },0)$ where $m<\mathsf{\ell }\le k$, then $h\left(e_2\right)=(m+\mathsf{\ell },m)$, hence the $\left({\displaystyle \genfrac{}{}{0pt}{}{k+1}{2}}\right)$ homotheties sends $e_1$ to distinct points that are not in $\mathsf{\Phi }(2,k)$. The union of the images of $S_3$ under these homotheties is therefore the set

$$\Delta (3,k,1)=\{(i,j):0\le j\le i\le k\},$$

(29)

which has $\left({\displaystyle \genfrac{}{}{0pt}{}{k+2}{2}}\right)$ points in it. Hence, for all $k\ge 1$,

$$\begin{array}{c}\hfill \left|\Delta \left(3,k,1\right)\right|=\left({\displaystyle \genfrac{}{}{0pt}{}{k+2}{2}}\right).\end{array}$$

(30)

By (6) with $n=3$ and $m=1$, we get

$$\begin{array}{cc}\hfill \Delta \left(3,k,2\right)& =\Delta \left(3,k^{|\Delta \left(3,k,1\right)|},1\right)+\Delta \left(3,k,1\right).\hfill \end{array}$$

(31)

To get $\Delta \left(3,k,2\right)$ we must add all the vectors in $\Delta \left(3,k^{|\Delta \left(3,k,1\right)|},1\right)$ to all the vectors in $\Delta (3,k,1)$. To do this we let $T\left(\mathsf{\ell }\right)=\Delta (3,\mathsf{\ell },1)$ for every $\mathsf{\ell }\ge 1$ and prove that if $\mathsf{\ell }<m$ then

$$\begin{array}{cc}\hfill T\left(\mathsf{\ell }\right)+T\left(m\right)& =T(\mathsf{\ell }+m).\hfill \end{array}$$

(32)

Suppose $(i,j)\in T\left(\mathsf{\ell }\right)$ and $(i^{\prime },j^{\prime })\in T\left(m\right)$, i.e., $0\le j\le i\le \mathsf{\ell }$ and $0\le j^{\prime }\le i^{\prime }\le m$. Then $(i,j)+(i^{\prime },j^{\prime })=(i+i^{\prime },j+j^{\prime })\in T(\mathsf{\ell }+m)$ because the hypothesized inequalities imply $0\le j+j^{\prime }\le i+i^{\prime }\le \mathsf{\ell }+m$. This shows $T\left(\mathsf{\ell }\right)+T\left(m\right)\subseteq T(\mathsf{\ell }+m)$. For the converse, we have assumed $\mathsf{\ell }\le m$, so $T\left(\mathsf{\ell }\right)\subseteq T\left(m\right)$. Suppose $(i,j)\in T(\mathsf{\ell }+m)$, i.e., $0\le j\le i\le \mathsf{\ell }+m$. If $(i,j)\in T\left(\mathsf{\ell }\right)$ then $(i,j)=(i,j)+(0,0)\in T\left(\mathsf{\ell }\right)+T\left(m\right)$ because $(0,0)\in T\left(m\right)$. Therefore, assume $(i,j)\notin T\left(\mathsf{\ell }\right)$. Since $j\le i$, this assumption implies that $\mathsf{\ell }<i$, hence $0<i-\mathsf{\ell }$. Consider two cases. If $j\le \mathsf{\ell }$ then $(i,j)=(\mathsf{\ell },j)+(i-\mathsf{\ell },0)\in T\left(\mathsf{\ell }\right)+T\left(m\right)$. If $\mathsf{\ell }<j$ then $(i,j)=(\mathsf{\ell },\mathsf{\ell })+(i-\mathsf{\ell },j-\mathsf{\ell })\in T\left(\mathsf{\ell }\right)+T\left(m\right)$. This proves (32).

It follows from (31) and (32) that

$$\begin{array}{cc}\hfill \Delta \left(3,k,2\right)& =\Delta (3,k^{|\Delta \left(3,k,1\right)|}+k,1),\hfill \end{array}$$

(33)

and from (30) and (33) that

$$\begin{array}{cc}\hfill \left|\Delta \left(3,k,2\right)\right|& =\left({\displaystyle \genfrac{}{}{0pt}{}{k^{|\Delta \left(3,k,1\right)|}+k+2}{2}}\right).\hfill \end{array}$$

(34)

If we take one more step, then

$$\begin{array}{cccc}\hfill \Delta \left(3,k,3\right)& =\Delta \left(3,k^{|\Delta \left(3,k,2\right)|},1\right)+\Delta \left(3,k,2\right)\hfill & \hfill & \mathrm{by}\left(6\right)\mathrm{with}m=2\hfill \\ & =\Delta \left(3,k^{|\Delta \left(3,k,2\right)|},1\right)+\Delta (3,k^{|\Delta \left(3,k,1\right)|}+k,1)\hfill & \hfill & \mathrm{by}\left(33\right)\hfill \\ & =\Delta \left(3,k^{|\Delta \left(3,k,2\right)|}+k^{|\Delta \left(3,k,1\right)|}+k,1\right)\hfill & \hfill & \mathrm{by}(32),\hfill \end{array}$$

so by (30) we have

$$\begin{array}{cc}\hfill \left|\Delta \left(3,k,3\right)\right|& =\left({\displaystyle \genfrac{}{}{0pt}{}{k^{|\Delta \left(3,k,2\right)|}+k^{|\Delta \left(3,k,1\right)|}+k+2}{2}}\right).\hfill \end{array}$$

(35)

Ignoring lower order terms and constants, we can see from (30), (34) and (35) that as m increases, $\Delta \left(3,k,m\right)$ grows like an exponential tower of k’s, where the height of the tower is m. To get to $\left|\mathsf{\Phi }\right(3,k\left)\right|$ we need an exponential tower of k’s of height k. For example, to compute $\left|\mathsf{\Phi }(3,3)\right|=\left|\Delta \left(3,3,3\right)\right|$, let $k=3$ in (30), (34) and (35) to obtain

$$\begin{array}{cc}& \left|\Delta \left(3,3,1\right)\right|=10,\hfill \\ & \left|\Delta \left(3,3,2\right)\right|=\mathrm{1,743,657,931},\hfill \\ \hfill \left|\mathsf{\Phi }(3,3)\right|=& \left|\Delta \left(3,3,3\right)\right|=\left({\displaystyle \genfrac{}{}{0pt}{}{3^{\mathrm{1,743,657,931}}+3^{10}+3+2}{2}}\right)=239\cdots ,\hfill \end{array}$$

an integer with 1,663,872,520 digits [6]. When the number of points is larger than 3, the functions involved will contain towers of towers of exponents.

The numbers involved are clearly much larger than necessary. To illustrate this in the case at hand, recall that, by Theorem 2 with $n=3$ and $k=2$, there is a monochromatic homothetic image of $S_3$ in $\mathsf{\Phi }(3,2)$ whenever $\mathsf{\Phi }(3,2)$ is colored with two colors. Recall that the a priori upper bound on $\left|\mathsf{\Phi }\right(3,2\left)\right|$ derived earlier was 12,870, but for the particular choice of $S_3$ considered here, the actual value that arises from the proof is considerably smaller. By (7), (30) and (34) with $n=3$ and $k=2$, we get $\left|\Delta \left(3,2,1\right)\right|=\left({\displaystyle \genfrac{}{}{0pt}{}{2+2}{2}}\right)=6$ and

$$\begin{array}{cc}\hfill \left|\mathsf{\Phi }(3,2)\right|=\left|\Delta \left(3,2,2\right)\right|& =\left({\displaystyle \genfrac{}{}{0pt}{}{2^{|\Delta \left(3,2,1\right)|}+2+2}{2}}\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{2^6+2+2}{2}}\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{68}{2}}\right)=2278.\hfill \end{array}$$

Thus, by Theorem 2, if the set $\mathsf{\Phi }(3,2)$, which contains 2278 points, is colored with two colors, then it contains a monochromatic homothetic image of $S_3$. However, $\Delta \left(3,4,1\right)$ already has this property and it contains only 15 points by (29) and (30). The proof of this observation, stated next, is an easy if somewhat tedious exercise.

Theorem 3.

Let $S_3=\{e_0,e_1,e_2\}$ where $e_0=(0,0)$, $e_1=(1,0)$, and $e_2=(1,1)$. Let $Y=\{(i,j):0\le j\le i\le 4\}$. If Y is colored with two colors, then Y contains a monochromatic homothetic image of $S_3$.

Gallai’s theorem may conceivably need 12,870 points for something that in this case requires no more than 2278 points but which actually needs only 15 points. Gallai’s theorem, like Ramsey’s theorem, guarantees things eventually happen that actually happen much earlier, but figuring out when can be very challenging.

10. Concluding Remarks

Gallai’s theorem, a central result in Euclidean Ramsey theory, can be simply deduced as a consequence of the Hales–Jewett theorem [7], but the purpose here has been a detailed analysis of Witt’s proof. The key steps and constructions have been isolated in the recursive definitions of the sets $\mathsf{\Phi }(n,k)$ and $\Delta (n,k,m)$ and the statements $A(n,k)$ and $D(n,k,m)$. These may make Gallai’s theorem more accessible and could form the basis of further formalization or generalization.