$R^n$ satisfies the second axiom of countability

Question

I proved it but I'm not sure if it's right. The definition of the basis for the euclidean topology on $R^n$ is $\beta = \{<x_1,x_2,...,x_n>\in R^n\, : \, a_i<x_i<b_i,\, i = 1,2,...,n\}$. So my proof is:

Put the basis $\beta' = \{<x_1,x_2,...,x_n>\, : \, n<x_i<n+1,\, n\in Z\, and\, i = 1,2,...,n\}$, then $R^n$ satisfies the second axiom of countability.

Because $\beta'$ contains a countable number of elements. I think $\beta'$ is not a basis, what's wrong?

Answer 1

The set of real number $\mathbb{R}$ with the standard topology is second countable. For example, $$ \mathscr{B}=\{ (a,b) \subset \mathbb{R} \mid a<b,\ a,b\in\mathbb{Q} \} $$ is a countable basis of $\mathbb{R}$, where $\mathbb{Q}$ is the set of rational numbers.

If you agree with the fact above, it is easy to find a countable basis for $\mathbb{R}^n$, for example, $$ \mathscr{B}'=\{ (a_1,b_1)\times(a_2,b_2)\times\dotsb\times(a_n,b_n) \subset \mathbb{R}^n \mid a_i<b_i,\ a_i,b_i\in\mathbb{Q} \text{ for all $1\leq i\leq n$} \} $$