Prove that every separable metric space (say X) has a countable base. (Hint: take all neighborhoods with rational radius and center in some countable dense subset of X).
My question is: Is it necessary to take a rational radius? I mean since it is given that X is separable so it has some countable dense set. For creating base we'll use the said countable dense subset and we can consider a ball with the center from the subset, so the no. of balls will still be countable. I don't see why do we need a rational radius. Please Clarify this.