Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 2 / 20F (Linear Analysis)

Because $Y$ is closed, it is a nowhere dense set exactly when its interior is empty. If a ball about $y\in Y$ lay inside $Y$, translating it by $-y$ would put a ball about $0$ inside $Y$. Every vector of $X$ has a small scalar multiple in that ball, and closure under scalar multiplication would force $Y=X$. This contradiction proves a proper closed linear subspace is nowhere dense.

One form of the Baire category theorem is: a nonempty complete metric space cannot be a countable union of closed nowhere dense subsets. Equivalently, a countable intersection of open dense subsets is dense. A Banach space is a complete metric space for its norm metric, so this form applies directly to the next part.

Suppose $(x_k{)}_{k\ge 1}$ were a countable Hamel basis. Put $E_n=\mathrm{span}(x_1,\dots ,x_n)$. Each $E_n$ is a finite-dimensional, hence closed, linear subspace of $X$. It is proper because $X$ is infinite dimensional, and therefore a nowhere dense set. A Hamel basis means that every vector is a finite linear combination, so $X={\bigcup }_n{E}_n$. This contradicts the Baire category theorem for the Banach space $X$. Hence an infinite-dimensional Banach space has no countable Hamel basis. The same proof excludes any countable algebraic spanning set.