As part of the recent series on how I approach maths problems, I give another one here (question 14 on the Maths Tripos IB 2007 paper 4). The question is:
Show that a compact metric space has a countable dense subset.
This is intuitively clear if we go by our favourite examples of metric spaces (namely \(\mathbb{R}^n\), the discrete metric and the indiscrete metric). Indeed, in \(\mathbb{R}^n\), which isn’t even compact, we have the rationals (so the theorem doesn’t give a necessary condition, only a sufficient one); in the indiscrete metric, any singleton \({x }\) is dense (since the only closed non-empty set is the whole space); in the discrete metric, where every set is open, we can’t possibly be compact unless the space is finite, so that’s why the theorem doesn’t hold for a topology with so many sets.