Is the Collection of Rational Balls a Basis for the Euclidean Topology on R^n?

[littleHilbert]

Hello,

there is a basic lemma in topology, saying that:
Let X be a topological space, and B is a collection of open subsets of X. If every open subset of X satisfies the basis criterion with respect to B (in the sense, that every element x of an open set O is in a basis open set S, contained in O), then B is a basis for the topology of X.

With this lemma at hand, it is asked to show that the collection:
\mathcal{B}=\{\text{the set of balls}\, B_r(x)\, \text{with rational radius, where x has rational coordinates}\} is a basis for the Euclidean topology on R^n.

Take any set O, which is open in the Euclidean metric space sense, i.e. any point in O has a ball of radius \varepsilon around it, which is contained in O.

1. If x has rational coordinates, take a ball, whose radius is a rational number smaller than the given \varepsilon.
2. If x has irrational coordinates, we know that the set of points with irrational coordinates is dense in R^n. Hence any such x is between some points with rational coordinates, say a and b. Then it is in the ball around a with radius r=b-a, which is from \mathcal{B} and by 1. contained in O.

Is this explanation correct?

 

[g_edgar]

It is perhaps best not to use "between" in dimensions higher than 1.
And certainly r=a-b to get the radius is an error when a,b are multi-dimensional.

 

[littleHilbert]

Ups, of course I meant r=|b-a|. Sorry for that dumb misprint! Thanks!