About dense subsets of Tychonoff products of discrete spaces (Q2400871)

It is well known that a product of no more than continuum many separable spaces is again separable. To prove this it suffices to prove that the particular power~\(\mathbb{N}^\mathbb{R}\), where \(\mathbb{N}\)~has the discrete topology, is separable. Many constructions produce a sequence \(\langle x_n:n\in\mathbb{N}\rangle\) of points in the product via a family \(\{f_r:r\in\mathbb{R}\}\) of functions from \(\mathbb{N}\) to itself. If one thinks of this family as an \(\mathbb{R}\times\mathbb{N}\)-matrix then one obtains \(\langle x_n:n\in\mathbb{N}\rangle\) by transposing this matrix: \(x_n(r)=f_r(n)\). The properties of the family of functions influence the properties of the corresponding countable dense set. In the present paper the author shows that one can create a countable dense set, \(Q\), that can be written as a union of disjoint finite sets \(Q_k\) such that: (1)~if \(F\subseteq Q\)~is such that \(|F\cap Q_k|\leq1\) for all \(k\) then \(F\)~is closed and discrete in the full product and (2)~if \(C\subseteq\mathbb{N}\)~is infinite and \(F\subseteq Q\) is such that \(|F\cap Q_k|\leq m\) for some fixed~\(m\) and all~\(k\) then \(\bigcup_{k\in C}Q_k\setminus F\) is dense in the product. As a consequence no sequence in~\(Q\) converges in \(\mathbb{N}^\mathbb{R}\).