We prove that for any sequence of binary alphabets $\mathcal{ A}_1,\mathcal{ A}_2,\dots$, there exists a cube-free word $c_1c_2\dots$ so that $c_1\in\mathcal{ A}_1,c_2\in\mathcal{ A}_2,\dots$. In particular, for every $n$, there are at least $1.35^n$ cube-free words in $\mathcal{ A}_1\times\mathcal{ A}_2\times\dots\times \mathcal{ A}_n$. We also prove that if the list of alphabets is computable then one of these words is computable and its $n$th letter can be computed in time polynomial in $n$.
Categories: math.CO