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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 :)

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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
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up vote 11 down vote accepted

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 $\{\sum_{k=1}^\infty t^ke_k:0\lt t\lt 1\}$.

The same fact extends to Banach spaces, but there orthogonality cannot be used to write so short of a proof. A proof is given in this short article by Lacey.

To just see that the dimension cannot be countable, you could use Baire's theorem.

For more on this in the Hilbert space case, see Problem 7 of Halmos's Hilbert space problem book. (It is assumed there that the Hilbert spaces are complex instead of real, but this does not affect your question.)

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