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.


1 Answer 1


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 $n = 1, 2, \ldots$.

  • $\begingroup$ Suppose I've created a ball and it's an element of a base. so while selecting the radius can't we just take a fixed irrational number which would satisfy the condition of it becoming an element of a base $\endgroup$
    – Lucas
    Commented Aug 19, 2020 at 15:01
  • $\begingroup$ We are trying to construct a countable base. It needs to contain arbitrarily small neighbourhoods of each point in $X$. $\endgroup$
    – Rob Arthan
    Commented Aug 19, 2020 at 15:06
  • $\begingroup$ Okay, got it. Thanks $\endgroup$
    – Lucas
    Commented Aug 19, 2020 at 15:18

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