Q: I can't get the sketch of the proof Now since an arbitrary element of $X$ having the expansion $\sum_{n=1}^\infty \alpha_n b_n$ in terms of $(b_n)_{n \in N}$ can be written as $\lim _{N \rightarrow \infty} \sum_{n=1}^N \alpha_n b_n,$ it follows that $X = \overline{span(\{x_n : n \in N\})}$ so $span(\{x_n : n \in N\})$ is already dense in X. what we prove then?
Also, i see that proof for the same theorem
Q: why because $\overline{span(\{x_n\})} = X$ then it is suffices to prove that $A$ is a dense subset of $span(\{x_n\})$