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.