Colouring finite products (Q2077893)

Let \(X_1,\dots,X_n\) be a finite sequence of countable infinite sets. Then, there is a colouring of the product \(\prod_{i=1}^n X_i\) with \(n!\) colours such that for each family \(\{ Y_i :i<n\}\) of infinite subsets \(Y_i \subseteq X_i\), each of the \(n!\) colours shows up in the subproduct \(\prod_{i=1}^n Y_i\).\par Here, the author investigates this phenomenon in case when some of the sets \(X_i\) are uncountable. He shows:\par Theorem 1.1: Suppose \(X_1,\dots,X_n\) are infinite sets with at least one of them uncountable. Then for each finite colouring of \(\prod_{i=1}^n X_i\) there exists infinite sets \(Y_i \subseteq X_i\) such that \(\prod_{i=1}^n Y_i\) uses no more that \((n-1)!\) colours.\par Theorem 1.2: Let \(\Omega\) denote the set of countable ordinals. Then for each \(n<\omega\) there is a finite colouring of \(\Omega^n\) which has at least \((n-1)!\) colours on any subproduct \(\prod_{i=1}^n Y_i\) with \(Y_i \subseteq \Omega\) infinite.\par The first theorem can be seen as an interpretation of a result of R. Laver. For the proof of the two theorems, the author uses \(\rho\)-functions which he studied in detail in \textit{S. Todorcevic} [Walks on ordinals and their characteristics. Basel: Birkhäuser (2007; Zbl 1148.03004)].