Show that if \(X\) is a countable product of spaces having countable dense subsets, then \(X\) has a countable dense subset.
Solution
Let \[ X=\prod_{n=1}^\infty X_n, \] and for each \(n\) choose a countable dense set \(D_n\subset X_n\). Choose a basepoint \(a_n\in D_n\). Let \(D\subset X\) consist of all sequences \(x=(x_n)\) such that \(x_n\in D_n\) for every \(n\) and \(x_n=a_n\) for all but finitely many \(n\).
The set \(D\) is countable: it is a countable union, over finite subsets \(F\subset\mathbb Z_+\), of the countable products \(\prod_{n\in F}D_n\).
It is dense. A basic nonempty open set in the product topology restricts only finitely many coordinates, say \(n\in F\), with \(x_n\in U_n\). Since \(D_n\) is dense, choose \(d_n\in D_n\cap U_n\) for \(n\in F\), and set all other coordinates equal to \(a_n\). The resulting point lies in \(D\) and in the basic open set. Hence \(D\) is countable dense.