# Clarification on Every separable metric space has a countable base

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.

Taking just one ball round each point in the dense subset would not give a base. If you allowed the radius to be arbitrary, then there would be uncountably many neighbourhoods in general. Taking the balls with rational radius gives a countable base. You could use any countable set of non-zero radii provided that, for every $$\epsilon > 0$$, you include a ball with radius less than $$\epsilon$$. E.g., you could just take the balls with radius, $$1/n$$ for $$n = 1, 2, \ldots$$.
• We are trying to construct a countable base. It needs to contain arbitrarily small neighbourhoods of each point in $X$. – Rob Arthan Aug 19 '20 at 15:06