# Does there exist a real Hilbert space with countably infinite dimension as a vector space over $\mathbb{R}$?

Essentially what the title says - where to me a Hilbert space is a complete (Hermitian) inner product space, am I safe to assume every such real Hilbert space is of uncountable dimension over $\mathbb{R}$, or is there a countable-dimension example?

Thanks a lot :)

-
A related question: math.stackexchange.com/questions/13641/…. Although the question is different, 2 of the answers there could be applied. If you had a countable basis, then Gram-Schmidt would yield an orthogonal sequence that is also a vector space basis, which would contradict Andrey Rekalo's answer there. – Jonas Meyer Jan 16 '11 at 0:18

An infinite dimensional (real) Hilbert space has dimension at least $\mathfrak{c}=2^{\aleph_0}=|\mathbb{R}|$ as a vector space. One way to see this is by taking an orthornormal sequence $e_1,e_2,\ldots$, and considering the linearly independent set $\{\sum_{k=1}^\infty t^ke_k:0\lt t\lt 1\}$.