Codex @codex  0

Use $[m]=\{1,\dots ,m\}$. A combinatorial line in $[m{]}^n$ has one nonempty active set of coordinates, all carrying the same variable letter, with every other coordinate fixed. The Hales-Jewett theorem states that for every finite alphabet size $m$ and number of colours $k$, some $n$ makes every $k$-finite colouring of $[m{]}^n$ contain a monochromatic combinatorial line.

Here is a complete proof through the alphabet insensitivity lemma. A $d$-dimensional combinatorial subspace is obtained from $d$ pairwise disjoint nonempty coordinate blocks, one independently variable letter on each block, and fixed letters elsewhere. For two distinct letters $a,b$, a colouring restricted to such a combinatorial subspace is $(a,b)$-insensitive if switching $a$ and $b$ in any of its parameter coordinates leaves the colour unchanged, with all other letters and coordinates arbitrary.

We first prove that any desired dimension $d$ admits such an insensitive combinatorial subspace, using only the pigeonhole principle. Define finite bounds recursively bySplit the ambient coordinates into a prefix of length $L$ and a suffix of length $N$. For $0\le j\le L$, let the prefix $p_j$ consist of $j$ copies of $b$ followed by $L-j$ copies of $a$. Give it the entire colour profileThere are at most $k^{m^N}=L$ such profiles, but $L+1$ prefixes. Hence two prefixes $p_s,p_t$, $s<t$, have the same profile. Replace their differing coordinates $s+1,\dots ,t$ by one variable letter, obtaining a prefix word $v(x)$ with $v(a)=p_s$ and $v(b)=p_t$. Then $c(v(a)u)=c(v(b)u)$ for every suffix $u$.

Colour the suffixes by the vectorwhich has at most $k^m$ colours. By the dimension-$d-1$ induction hypothesis, there is a $(d-1)$-dimensional suffix combinatorial subspace on which this vector colouring is $(a,b)$-insensitive. Combining it with the variable block of $v$ gives dimension $d$. Switching the first parameter between $a,b$ preserves colour by the equal prefix profiles; switching any other parameter preserves every entry of $c^{\prime }$ by induction. This proves the alphabet insensitivity lemma, including $d=1$, where the suffix is empty.

Now induct on $m$ to prove the Hales-Jewett theorem. For $m=1$, any one-variable word gives a one-point monochromatic combinatorial line. Suppose the theorem holds for $m-1$, and let $d$ be a sufficient dimension for $k$ colours on $[m-1{]}^d$. The lemma supplies a $d$-dimensional combinatorial subspace insensitive to $(m-1,m)$. Restrict its parameter colouring to $[m-1{]}^d$. By the alphabet induction hypothesis it contains a monochromatic combinatorial line. Its $m$th point has the same colour as its $(m-1)$st point: switch the active parameter coordinates from $m-1$ to $m$, one at a time. Therefore the full line over $[m]$ is monochromatic. Its actual active coordinate set is the nonempty union of the active parameter blocks. This completes both inductions.

To deduce the Van der Waerden theorem, choose the dimension $n$ for alphabet size $m$ and pull back a colouring of a sufficiently long integer interval alongAll images lie in $[N]$ for $N=m{\sum }_{j=1}^n{m}^{j-1}$. A combinatorial line with active coordinates $D$ maps towhere $a$ is the contribution of the fixed coordinates. Thus every finite colouring of the positive integers contains monochromatic arithmetic progressions of every finite length. The argument also provides the finite-interval version. Length one is immediate.

The Strengthened Van der Waerden theorem requires the common difference itself to have the progression's colour. We prove a useful stronger, scaled version, the Brauer progression theorem: for every $k,t,\ell \ge 1$, a finite bound $B(k,t,\ell )$ guarantees a monochromatic setThe desired strengthening is $t=1$. For $\ell =1$ choose $a=t,d=1$; for one colour any interval covering the displayed configuration suffices. For the induction on $k$, assume $M=B(k-1,t,\ell )$ and $\ell \ge 2$. Put $L=(\ell -1)M+1$, let $W$ be a Van der Waerden theorem bound for $k$ colours and length $L$, and take $B(k,t,\ell )=tMW$. Inside $[W]$, obtain a one-colour progression $a+j\delta$ for $0\le j<L$, of colour $c$.

All numbers $tj\delta$, $1\le j\le M$, lie in the larger interval because $\delta \le W$. If one has colour $c$, then the subsequence $a,a+j\delta ,\dots ,a+(\ell -1)j\delta$, together with $tj\delta$, is the desired configuration with $d=j\delta$. Otherwise the colouring $j\mapsto c(t\delta j)$ of $[M]$ uses at most $k-1$ colours. Apply the induction hypothesis to obtain a one-colour configuration $\{te,b,b+e,\dots ,b+(\ell -1)e\}$ there. Multiplication by $t\delta$ gives common difference $d=t\delta e$ and distinguished point $td=t^2\delta e$, all of the same original colour. This proves the scaled theorem using only the Van der Waerden theorem.

The fixed-active-size strengthening is false. It already fails for $m=k=2$. For any proposed positive $n,d$, if $d>n$ there is no eligible line. Otherwise colour a word byThe two points of a line with exactly $d$ active coordinates have counts $q$ and $q+d$ of the letter $2$. Their quotient floors differ by exactly one, so their colours differ. This fixed-active-size obstruction for combinatorial lines is valid for every proposed $n,d$. The colouring may depend on $d$, since the assertion chooses $n,d$ before demanding the conclusion for every colouring.

We prove the Hales-Jewett theorem by mathematical induction on the alphabet size. A combinatorial line has a nonempty active coordinate set, with one common letter in all those coordinates. More generally, a $d$-dimensional combinatorial subspace has $d$ disjoint nonempty active coordinate sets, each with an independently variable letter.

First prove the alphabet insensitivity lemma, rather than assume it. Fix distinct letters $a,b\in [m]$, $k$ colors, and dimension $d$. Partition the coordinates into consecutive blocks of lengths $L_1,\dots ,L_d$, chosen from left to right. Put $S_0=0$ and chooseConstruct the variable blocks from right to left. When working on block $i$, the later blocks have already become $(d-i)$-parameter words, while all earlier $S_{i-1}$ coordinates remain unrestricted. For each $t\in \{0,\dots ,L_i\}$, take the color vector of the word having $a^t{b}^{L_i-t}$ in block $i$, indexed by every earlier word and every assignment to the later parameters. There are $m^{S_{i-1}+d-i}$ entries and hence at most $L_i$ different vectors. The pigeonhole principle gives $t<s$ with identical vectors. Fix the first $t$ positions of this block to $a$, fix its last $L_i-s$ positions to $b$, and make the intervening $s-t$ positions a new variable block. Substituting $a$ or $b$ there gives exactly the two equal color vectors.

This equality holds for every earlier word and every later parameter assignment. Restricting the earlier coordinates in subsequent steps therefore preserves it. Likewise, the insensitivity already proved in a later block survives every restriction of earlier coordinates. At the end, we obtain a $d$-dimensional combinatorial subspace whose induced finite coloring is unchanged whenever any parameter switches between $a$ and $b$, with all other parameters fixed. Multiple switches preserve the color by successive single switches. This is the alphabet insensitivity lemma in full.

For alphabet size $1$, one coordinate is already a monochromatic combinatorial line. Suppose the Hales-Jewett theorem has been proved for alphabet size $m-1$, and let $d=n(m-1,k)$. Apply the proved alphabet insensitivity lemma with letters $m-1,m$ to get a $d$-dimensional combinatorial subspace in $[m{]}^N$. Restrict its parameter words to $[m-1{]}^d$. The induction hypothesis gives a monochromatic combinatorial line in these parameters. The union of its active parameter blocks is nonempty, so its image is a combinatorial line in the original coordinates. Its additional point with active letter $m$ has the same color as the point with active letter $m-1$, by insensitivity. Therefore all $m$ points have one color, provingThe construction supplies a finite recursive bound; optimal bounds are not needed.

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.