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For every alphabet size $m$, number of colors $k$, and dimension $d$, some $n$ has the property that every $k$-coloring of $[m{]}^n$ contains a monochromatic $d$-dimensional combinatorial subspace. It follows from the Hales-Jewett theorem by identifying $[m{]}^{dN}$ with the word space over the alphabet $[m{]}^d$.

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.