Let $E$ be an infinite dimensional Banach space and let $(u_n)_{n\ge1}$ be a sequence of elements in $E$. Let $F$ be the vector subspace of $E$ consisting of finite linear combinations of the $u_n$. Prove that we cannot have $F = E$.
Plus: Let $E$ be the Banach space of continuous functions on the compact interval $[a, b]$ with values in $\mathbb{R}$ with the norm $|| · ||_∞$. Give an example of a sequence $(u_n)_{n \ge 1}$ in $E$ such that the space $F$ defined above has its closure equal to $E$.
I do not know how to start, ideas and suggestions are appreciated, many thanks!
Suppose $F=E$. Then $E=\cup_n M_n$ where $M_n=span \{u_1,u_2,..,u_n\}$. Any finite dimensional subspace is closed and a proper subspace has no interior. Hence $M_n$'s are nowhere dense. So Baire's Theorem gives a contradiction.
For the second part consider the class of all polynomials with rational coefficients. This is a countbale set and we can write it as a sequence $(u_1,u_2,...)$. The fact that $F=E$ in this case follows easily from Weierstrass Approximation Theorem.