Denumerable dimensional tvs I am trying to prove the above. However, I even cannot understand why it is true. For example, $\ell^2$ is a complete normed space which has countable orthonormal basis and so countable dimensional. Am I right? Why this one is not a counter example for above? What am I missing?
I am really confused on this. I am not sure how to start to this proof. Any helps or hints will be appreciated.
Thanks a lot.
 A: There are two different notions of basis in use here: Hamel basis and Schauder basis.
Definition of Hamel basis implies that linear combinations of finitely many vectors from a basis span the whole space. This is not the case for a Schauder basis, which is defined in terms of countable linear combinations.
The theorem stated in your question concerns dimension defined in terms of cardinality of Hamel bases.
As for the example of $\ell^2$ it in fact has a (countable) Schauder basis. The theorem stated in the question implies that it does not have a countable Hamel basis.
As for the proof of the theorem, assume there exists an infinite-dimensional complete linear topological space with countable Hamel basis $\{ e_i \}_{i=1}^{\infty}$. Consider subspaces $V_k = lin\{e_1, e_2, \ldots, e_k\}$. Each of subspaces $V_k$ is closed and has empty interior. Hence by the Baire category theorem $V = \bigcup_{k=1}^{\infty} V_k$ has an empty interior too. On the other hand $V$ equals the whole space, hence we obtain a contradiction. $\square$
