Let \(X\) and \(Y\) be two topological spaces and let \(X\times Y\) be endowed with the product topology. Prove that if \(X\) and \(Y\) each have a countable dense subset, then so does \(X\times Y\).
Solution
Let \(D_X\subset X\) and \(D_Y\subset Y\) be countable dense subsets. Then \(D_X\times D_Y\) is countable.
Every nonempty basic open set in the product topology has the form \(U\times V\) with \(U\subset X\) and \(V\subset Y\) nonempty and open. Density gives points \(x\in U\cap D_X\) and \(y\in V\cap D_Y\). Hence \[ (x,y)\in (U\times V)\cap(D_X\times D_Y). \] Thus \(D_X\times D_Y\) meets every nonempty basic open set and is therefore dense in \(X\times Y\). So \(X\times Y\) has a countable dense subset.