Recall the Hales-Jewett Theorem:



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> **HJT**: Given a finite alphabet $A$ and some $r \in \mathbb{N}$, there is some $H \in \mathbb{N}$ such that whenever $A^H$, the set of all length-$H$ words from $A$, is $r$-colored, there is a monochromatic combinatorial line.
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[A *combinatorial line* is defined as follows. Fix some $x \notin A$. A *variable word* $w$ is a length-$H$ word of members of $A \cup \{x\}$, with $x$ appearing at least once. For $a \in A$, $w(a)$ denotes the result of substituting $a$ for $x$ in $w$. A combinatorial line is any set of the form $\{w(a):a \in A\}$, where $w$ is a variable word. Notice that this definition makes sense even when $A$ or $H$ is allowed to be infinite.]


My question arises simply from wondering *what happens if we let $A$ be infinite?* If there is any hope of a positive result, we must be willing to let $H$ be infinite too. So I am asking about the following statement:



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> **HJT($\kappa$)**: Given a set $A$ with $|A| = \kappa$ and $r \in \mathbb{N}$, there is an infinite cardinal $\lambda$ such that whenever the set $A^\lambda$ is $r$-colored, there is a monochromatic combinatorial line.
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Using a result of William Weiss, I can prove that HJT($2^{\aleph\_0}$) implies the existence of measurable cardinals. But that is about all I know.



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> **Question 1**: Is HJT($2^{\aleph\_0}$) consistent with ZFC plus some large cardinal assumption?
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> **Question 2**: What about HJT($\kappa$) more generally?
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The second question is admittedly broad, but that's because I know so little. Any results, pointers, or consistency proofs would be welcome.



The Hales-Jewett Theorem for an infinite alphabet
I returned to this problem recently, and was finally able to answer my own question (though not in the way I'd hoped):



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> **Theorem:** The principle HJT($\kappa$) is false for every infinite cardinal $\kappa$.
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As a consolation prize, we can at least get:



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> **Theorem:** Suppose $A$ is countable and $A^\omega$ is colored with a Borel function. Then there is a monochromatic combinatorial line.
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> **Theorem:** If ZF is consistent, then so is ZF+DC+HJT($\aleph\_0$).
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In other words, the Axiom of Choice can be used to obtain a counterexample to HJT($\kappa$) for any infinite $\kappa$. On the other hand, some form of Choice is needed, because it is consistent with ZF+DC that HJT($\aleph\_0$) holds.


To prove the first theorem, it suffices to show that HJT($\aleph\_0$) is false, since HJT($\kappa$) implies HJT($\aleph\_0$) for all infinite $\kappa$. Let $A = \omega$ and let $\lambda$ be any cardinal. We must exhibit a $2$-coloring of $\omega^\lambda$ with no monochromatic combinatorial lines.


Define an equivalence relation on elements of $\omega^\lambda$ as follows:
$$f \sim g \ \Leftrightarrow \ \{|f(\alpha) - g(\alpha)| : \alpha \in \lambda\} \text{ is bounded.}$$
If $f \sim g$, then define the *distance* from $f$ to $g$ to be $\max \{|f(\alpha) - g(\alpha)| : \alpha \in \lambda\}$.


Using the Axiom of Choice, pick a representative element from each $\sim$-equivalence class. Given $f \in \omega^\lambda$, determine the distance from $f$ to the chosen representative of its $\sim$-equivalence class. Color $f$ red if this distance is even, and otherwise color $f$ blue.


Now consider any combinatorial line in $\omega^\lambda$; it is determined by some variable word $w$. Notice that the combinatorial line is contained entirely within a single $\sim$-equivalence class; let $f$ be the chosen representative of this class. Let $d$ be the distance from $w(0)$ to $f$. Thus wherever $w$ has its variable, $f$ is at most $d$. It follows that, for $n \geq 2d$, the distance from $f$ to $w(n)$ is one less than the distance from $f$ to $w(n+1)$. In other words, when $n$ is sufficiently large the distance from $f$ to $w(n)$ depends only on whether $n$ is even or odd. Thus our monochromatic line contains both red and blue points.


To prove the second and third theorems, it suffices to take $A = \omega$ and to prove that if $\omega^\omega$ is partitioned into two sets, say $R$ and $B$, each of which has the [property of Baire](https://en.wikipedia.org/wiki/Property_of_Baire), then there is a monochromatic combinatorial line. (Note that this is HJF($\aleph\_0$) for two colors only, but any finite number of colors follows by induction. For the second theorem, note that every Borel set has the property of Baire. For the third, it is known that ZF+DC+"every subset of $\omega^\omega$ has the property of Baire" is consistent relative to ZF.)


$R$ and $B$ cannot both be meagre; let us assume that $R$ is not. Since we assume that $R$ has the property of Baire, we may find a clopen subset $[s]$ of $\omega^\omega$ such that $B \cap [s]$ is meagre. In particular, $B \cap [s^\frown n]$ is meagre for every $n \in \omega$; this means that, for every $n \in \omega$, there is meagre set of $\gamma \in \omega^\omega$ such that $[s^\frown n^\frown \gamma]$ is colored blue. As a countable union of meagre sets is meagre (this is a theorem of ZF+DC), there must be some $\gamma \in \omega^\omega$ such that $[s^\frown n^\frown \gamma]$ is red for every $n \in \mathbb N$.


Let $w$ be the variable word $s^\frown x^\frown \gamma$. The conclusion of the previous paragraph states that the combinatorial line determined by $w$ is monochromatic (in red).


