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Let $N$ be a word length supplied by the Hales-Jewett theorem for alphabet $[m]$ and two colors. For $n\ge 2N$, embed $[m{]}^N$ into $[m{]}^n$ by the coordinate duplication for combinatorial lines mapPull back any two-color finite coloring to this word space. The Hales-Jewett theorem gives a monochromatic combinatorial line with nonempty active set $S$. Its image is a combinatorial line whose active set isand hence has size $2|S|$, a positive even integer. ThereforeThe empty set's presence in the family causes no difficulty: the line just constructed has at least two active coordinates.

Hales–Jewett and the arithmetic-progression consequence. For a finite alphabet $[m]$ and a positive integer $r$, the Hales-Jewett theorem gives an integer $N$ such that every $r$-color finite coloring of $[m{]}^N$ contains a monochromatic combinatorial line. A combinatorial line has a nonempty active coordinate set: all those coordinates receive the same freely varying letter, and the remaining coordinates are fixed.

We first prove the alphabet insensitivity lemma. Fix two letters $a,b\in [m]$, with $m\ge 2$, and a desired dimension $d$. There is a combinatorial subspace of dimension $d$ on which changing $a$ to $b$ in any variable coordinate, with all other variable coordinates arbitrary, preserves the color.

For the construction, divide the coordinates into $d$ consecutive blocks. Choose their lengths from left to right byProcess the blocks from right to left. When processing block $i$, the later blocks already form a $(d-i)$-dimensional combinatorial subspace. Color each word in block $i$ by the vector of its original colors for every prefix word of length $P_i$ and every assignment to those later variables. There are $m^{P_i+d-i}$ contexts, and at most $r^{m^{P_i+d-i}}=L_i$ such color vectors.

The $L_i+1$ words $a^j{b}^{L_i-j}$, $0\le j\le L_i$, contain two, indexed by $u<v$, with the same vector by the pigeonhole principle. Fix the first $u$ coordinates to $a$, the coordinates after $v$ to $b$, and make coordinates $u+1,\dots ,v$ the active set of one new variable. Its values $a,b$ now have identical colors for every prefix and later-variable context. The active set is nonempty. Insensitivity already obtained in later blocks survives, because it held for every unrestricted earlier prefix. After all $d$ blocks have been processed, changing $a,b$ in any of the $d$ variables preserves the color. This proves the lemma.

Now induct on $m$. For $m=1$, a one-coordinate combinatorial line is automatically monochromatic. For $m\ge 2$, let $d$ be a length supplied by the theorem for alphabet $[m-1]$ and $r$ colors. Apply the alphabet insensitivity lemma for the letters $m-1,m$ to obtain a $d$-dimensional combinatorial subspace. Restrict its variables to $[m-1]$. Induction gives a monochromatic combinatorial line in those variables. Its values $1,\dots ,m-1$ have the same color, and its value $m$ has the same color as its value $m-1$ by successively switching the active variables. Lifting this line to the original coordinates proves

For the Van der Waerden theorem, fix a progression length $k\ge 2$ and use the alphabet $\{0,\dots ,k-1\}$. Pull back an $r$-color finite coloring of $\{1,\dots ,1+(k-1)N\}$ alongA combinatorial line with active set $S$ maps to the arithmetic progression $a,a+q,\dots ,a+(k-1)q$, where $a\ge 1$ and its common difference is $q=|S|>0$. The Hales-Jewett theorem makes this progression monochromatic. Length one is trivial, so every finite coloring of the positive integers contains monochromatic arithmetic progressions of every finite length.

The active-coordinate obstruction. Color a word red when its first coordinate is $1$, and blue otherwise. Any combinatorial line whose active coordinate set contains $1$ has a red word at variable value $1$ and a blue word at variable value $2$, since $m>1$. ThusHere adequacy is the adequate family of active coordinate sets property; an active coordinate set is always nonempty.

The Hales-Jewett theorem says that, for every finite alphabet $A$ and number of colors $r$, there is $N$ such that every $r$-coloring of $A^N$ contains a monochromatic combinatorial line.

We use the standard insensitivity lemma. Assuming the Hales-Jewett theorem for alphabets of size $m-1$, fix two letters $a,b$ in an $m$-letter alphabet. For every $r$ and $d$, there is $N$ such that each $r$-coloring of $A^N$ has a $d$-dimensional combinatorial subspace on which changing any selection of variable coordinates from $a$ to $b$, or from $b$ to $a$, leaves the color unchanged.

Here is the finite fusion proof of the lemma. Choose the sizes of $d$ successive coordinate blocks backwards. On the last block, omit $b$ and color a word over $A\setminus \{b\}$ by the vector of all colors obtained after filling the previously chosen blocks in every possible way and then replacing any chosen occurrences of $a$ by $b$. This is a finite derived coloring. The induction hypothesis supplies a combinatorial line on which that entire vector is constant. Treat its active coordinates as one new variable block and repeat. After $d$ repetitions, each replacement of $a$ by $b$ can be pushed into the last block where it was created, and equality of the derived color vectors shows that it does not alter the original color. This proves the insensitivity lemma.

Now induct on $m=|A|$. The case $m=1$ is immediate. For $m>1$, choose the target dimensions backwards and apply the insensitivity lemma successively to the pairsAt every step pass to the resulting nested combinatorial subspace, so the insensitivities already obtained are retained. On the final positive-dimensional subspace, replacing $a_m$ by any other letter does not change the color. Any two words can be connected by such replacements, so the whole subspace is monochromatic. It contains a combinatorial line, completing the induction and the proof.

The Gallai theorem for an integer lattice says that, for every finite $S\subseteq {\mathbb{N}}^d$ and every finite coloring of ${\mathbb{N}}^d$, there are $u\in {\mathbb{N}}^d$ and $q\ge 1$ such that the homothetic copyis monochromatic.

Write $S=\{s_1,\dots ,s_m\}$ and apply the Hales-Jewett theorem to the alphabet $[m]$. Color a word $w\in [m{]}^N$ by the color ofwhere a fixed positive vector $t$ keeps the image in ${\mathbb{N}}^d$ under either convention for the natural numbers. On a monochromatic combinatorial line, let $I$ be the active coordinate set. The fixed coordinates contribute a vector $u-t$, while the word whose active letter is $i$ maps toThese points form a monochromatic copy $u+|I|S$, proving Gallai's theorem.

Let $[m]$ be an alphabet. A combinatorial line in $[m{]}^N$ is obtained by choosing a nonempty set of active coordinates, fixing every other coordinate, and putting the same variable letter in all active coordinates. The Hales-Jewett theorem states that, for every pair of positive integers $m,k$, some $N$ makes every $k$-coloring of $[m{]}^N$ contain a monochromatic combinatorial line.

We prove it by mathematical induction on $m$. The case $m=1$ is immediate. Suppose the result is known for the alphabet $[m-1]$, with any finite number of colors. Order each line by its variable letter, and call its last point its focus. Say that $r$ lines are color-focused when they share a focus, each line with its focus removed is monochromatic, and those $r$ punctured lines have different colors.

We claim that for every $1\le r\le k$ there is $N_r$ such that every $k$-coloring of $[m{]}^{N_r}$ contains either a monochromatic line or $r$ color-focused lines. For $r=1$, restrict a coloring to $[m-1{]}^{N_1}$ with $N_1=HJ(m-1,k)$. A monochromatic line there becomes a punctured line over $[m]$ after adjoining the point obtained by putting $m$ in every active coordinate. It is either already monochromatic or is one color-focused line.

Assume $n=N_r$ works for $r$, and putView $[m{]}^{n+N}$ as $[m{]}^n\times [m{]}^N$. For $b\in [m-1{]}^N$, record the entire color patternThis is a coloring with at most $k^{m^n}$ colors, so the induction hypothesis on the alphabet supplies a line on which this pattern is constant. Adjoin its focus $b^{+}$, whose active coordinates contain the letter $m$, and write $L$ for the resulting line over $[m]$. Thus $c(a,b)$ is independent of $b\in L\setminus \{b^{+}\}$ for every $a$. It defines a $k$-coloring $c^{\prime }(a)$ of $[m{]}^n$.

If $c^{\prime }$ has a monochromatic line, fixing any $b\in L\setminus \{b^{+}\}$ gives one for $c$. Otherwise there are $r$ color-focused lines for $c^{\prime }$, with common focus $f$. Couple the active coordinates of each of those lines with the active coordinates of $L$; this produces $r$ punctured lines focused at $(f,b^{+})$. The line obtained by fixing the first component at $f$ and varying along $L$ supplies one more. Its punctured color differs from the preceding $r$ colors, since equality with one of them would make the corresponding $c^{\prime }$-line monochromatic. Hence there are $r+1$ color-focused lines. The claim follows by induction on $r$.

Take $r=k$. The common focus has one of the $k$ colors, so it completes the punctured line of that color to a monochromatic line. This proves the Hales-Jewett theorem.

A $d$-dimensional combinatorial subspace has $d$ disjoint nonempty active coordinate sets, one for each independent variable. The Extended Hales-Jewett theorem says that, given $m,k,d$, every $k$-coloring of $[m{]}^n$ contains such a monochromatic subspace when $n$ is large enough. To deduce it, chooseand identifyA monochromatic combinatorial line over the alphabet $[m{]}^d$ becomes a monochromatic $d$-parameter set over $[m]$: within every active block, its first coordinates form the first variable set, its second coordinates form the second, and so on. Thus $n=dN$ suffices.

The Hales-Jewett theorem states that for positive integers $m,k$ there is $N$ such that every $k$-coloring of the words $[m{]}^N$ contains a monochromatic combinatorial line.

Here is the standard focused-line proof. Induct on the alphabet size $m$, the case $m=1$ being immediate. Assume the result for $m-1$ and every number of colors. For $1\le s\le k$, prove inductively that some dimension has the following alternative: either there is a monochromatic combinatorial line, or there are $s$ color-focused lines, meaning that the lines without their common focus are monochromatic in $s$ distinct colors. For $s=1$, restrict to words on $[m-1]$ and use the induction hypothesis on $m$.

For the step from $s-1$ to $s$, let $n$ work for $s-1$ and view a longer word as $(a,b)\in [m{]}^n\times [m{]}^M$. Color each $b\in [m-1{]}^M$ by the complete patternwhich uses at most $k^{m^n}$ colors. Taking $M=HJ(m-1,k^{m^n})$ gives a line on which this entire pattern is constant. Append its missing $m$-letter endpoint. Applying the $s-1$ alternative to the induced coloring of the first block and joining the active coordinate sets produces either a monochromatic line or $s$ lines with one common focus and distinct colors. At $s=k$, the focus has one of the $k$ colors, so it completes the line carrying that color. This proves the theorem.

To deduce the Van der Waerden theorem, let $N=HJ(m,k)$ and color a word $(x_1,\dots ,x_N)\in [m{]}^N$ by the color of $x_1+\cdots +x_N$. On a combinatorial line with active set $I$, these sums area monochromatic arithmetic progression of length $m$.

Let $X$ be the vertex set of a regular polygon and let the cyclic rotation group $A$ act transitively on it. The finite invariant-colouring lemma says that for every number of colours there are positive weights $w_1,ldots,w_N$ with ${\sum }_i{w}_i^2=1$ such that every colouring of the weighted Cartesian powercontains a monochromatic setFor completeness, prove the lemma along a cyclic composition series for $A$. For a prime cyclic quotient, refine each colour to the finite vector of colours obtained by applying the quotient elements in every active coordinate. The Hales-Jewett theorem supplies a variable block on which this vector is constant. Assign that block squared weights summing to the squared weight of the coordinate it replaces. This makes the quotient orbit monochromatic without changing any orbit distance. Iterating through the cyclic factors proves the lemma for the finite cyclic group $A$.

Because rotations commute, for $a,b\in A$ each coordinate satisfiesfor a fixed vertex $v$. Hence the squared distance between the two corresponding product points isThe monochromatic orbit is therefore isometric to $X$. This proves that every regular polygon is a Euclidean Ramsey set.

Now consider the edge-colouring definition. If all pairwise distances in $X$ are equal, then $X$ is a regular simplex. Take a sufficiently large regular simplex $S$ of the same side length. The ordinary finite Ramsey theorem gives a monochromatic $|X|$-vertex complete subgraph, and every such vertex set is isometric to $X$. Thus $X$ is an edge Ramsey set.

Conversely, if $X$ has two distances, let $a$ and $b$ be its least and greatest distances. Colour every Euclidean edge by whether its length equals $a$. Every isometric copy of $X$ contains both an edge of length $a$ and one of length $b$, so no copy is monochromatic. Hence the edge Ramsey sets are exactly the equidistant finite sets.

Allowing similar copies does not change the answer. If $r=b/a>1$, colour an edge of length $d$ by the parity ofIn every similar copy, the images of a shortest and a longest edge have lengths $ta$ and $tb=rta$, whose displayed integers differ by one. They receive opposite colours. Thus no non-equidistant $X$ is edge Ramsey even up to similarity, while the regular-simplex argument already supplies an isometric copy.

For a finite pattern $\{v_1,\dots ,v_k\}$ in a real vector space, pull a finite colouring back to words through the displayed sum map. A combinatorial line with active coordinates J maps to $a+|J|\{v_1,\dots ,v_k\}$. The Hales-Jewett theorem therefore forces a monochromatic homothetic copy. Integer patterns remain in the integer lattice; choosing b sufficiently large gives the positive-lattice version.