# Infinite dimensional normed Banach space cannot have countable basis

It is said here that since an infinite-dimensional Banach space $M$ is meagre (it is contained in the countable union of nowhere dense closed subsets of itself), we reach a contradiction. However, I did not exactly understand what the contradiction is. Is it that the interior of a Banach space is necessarily non-empty, since it is open?

Just want to make sure I get it right.

• Baire category theorem states that any complete metric space is of second category, that is, comeagre. – Idonknow Nov 29 '17 at 10:36

Every finite dimensional subspace is closed (why?). Every proper subspace has empty interior (why?). So in an infinite dimensional space every finite dimensional subspace is nowhere dense. A countable dimensional subspace is a union of countably many finite dimensional subspaces (take span of the first $k$ basis vectors for each $k$—a nested union). So every countably-infinite dimensional subspace is first category. But the whole Banach space is second category by Baire’s theorem.