If $X$ is a banach space, then the span of countable independent vectors is never closed. I am trying to prove that statement in the title. I was trying to show that there is a sequence that converges to something outside the span, but i am struggling to do so. Let $S=\{v_k|k\geq 1\}$ be the set of linearly independent vectors. My idea was to consider $S'=\{a_k\}$ which is the normalized set of $S$ and then consider $\Sigma \frac{a_k}{k^2}$ which converges absolutely. Now i would like to show that the limit of this series doe snot belong to span$(S)$ but I am not sure how to do that.
Any help would be appriciated.
 A: Let $\{e_n\}_{n=1}^\infty$ be linearly independent and let $E$ be their span. Suppose $E$ is closed in $X$. Then $E$ is a Banach space, because it is a closed subspace of a Banach space. Now, for each $n\in\mathbb{N}$ let $E_n=span\{e_1,...,e_n\}$. Obviously $E=\cup_{n=1}^\infty E_n$. Since $E$ is a complete metric space it must be of second category (which means it isn't a countable union of nowhere dense sets), this follows from Baire's theorem. Hence there must be some $n\in\mathbb{N}$ such that $E_n$ is not nowhere dense in $E$. This means that the closure $\overline{E_n}$ contains an interior point. But note that $E_n$ is finite dimensional, hence closed. So $E_n=\overline{E_n}$ and we get that $E_n$ itself contains an interior point. But if a subspace contains a ball then it must be the whole space, hence $E=E_n$. But this is a contradiction since $E$ is not finite dimensional. (it contains an infinite sequence of linearly independent vectors) 
A: This is an easy consequence of Baire Category Theorem. If the span is closed it would be complete. If the countable set $x_1,x_2,..$ and $M_n$ is the span of $x_i: i \leq n$ then  the span of all the $x_i$'s is the union of the spaces $M_n$. Any proper subspace of $X$ has empty interior and any finite dimensional proper subspace is nowhere dense. Hence Baire Category Theorem is contradicted. 
