TITLE: No Hilbert space can have countable Hamel basis without using Baire's Category theorem
QUESTION [5 upvotes]: I have to prove that no Hilbert space can have countable Hamel basis just using the fact that any finite dimensional subspace is closed (more specifically without using Baire's theorem). I saw a paper by NAM-KIu TSING solving the same problem for Banach space. But, the proof is not much intuitive. Is it possible to give a easier proof for Hilbert space ?
Using proof by contradiction, the aim is to somehow find a Cauchy sequence and then use completeness to get a limit and show that cauchy sequence does not converge to that limit.
Thanks

REPLY [3 votes]: You don't even need to know that every finite-dimensional subspace is closed.  By Gram-Schmidt, given a countable Hamel basis, you can get a countable Hamel basis $\{e_n\}$ that is orthonormal.  But now if $(a_n)\in \ell^2$, the sequence of partial sums of $\sum a_n e_n$ is Cauchy and hence converges to some element $x$.  It is then easy to show that $\langle x,e_n\rangle=a_n$, and so if infinitely many $a_n$ are nonzero $x$ cannot be in the (Hamel) span of $\{e_n\}$.