Date: Nov 14, 2012 9:19 PM
Author: W^3
Subject: Re: Dimension of the space of real sequences

If R^N had a countable basis, then so would every subspace of R^N. In 
particular l^2 would have a countable basis, call it {v_1,_2, ...}.
Setting V_n = span {v_1, ..., v_n}, we then have l^2 = V_1 U V_2 U ...
But this violates Baire, as l^2 is complete (in its usual metric) and
each V_n is closed and nowhere dense in l^2.