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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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  • $\begingroup$ 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. $\endgroup$ – Jonas Meyer Jan 16 '11 at 0:18
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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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I think natural example of Hilbert space with a countable basis is a space of $L^2([-\pi,\pi])$ with a natural fourier set of basis functions ${e^{inx}, i \in -\infty, \ldots, \infty}$. See https://en.wikipedia.org/wiki/Fourier_series

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    $\begingroup$ That is not a vector space basis. It is not a maximal linearly independent set. It is an orthonormal basis, a maximal orthonormal set (with a scaling factor). That isn't asked here I think. $\endgroup$ – Jonas Meyer May 1 '17 at 23:52
  • $\begingroup$ When it is orthonormal, it is also linearly independent and it is maximal since if you add some function into it it will be linear combination of previous functions. Missing something? $\endgroup$ – VojtaK May 2 '17 at 15:25
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    $\begingroup$ If you add some function into an orthonormal basis no, it need not be a linear combination of previous functions. It will be a limit of linear combinations, with a series expansion, but it is not in the vector space span of the vectors in the ONB. E.g., $f(x) = x$ is not in the vector space span of the given ONB vectors. Linear independence of a set means linear independence of any finite subset. That is why a distinction is made between dimension "as a vector space" versus "Hilbert space dimension," the latter sometimes being used for the cardinality of an orthonormal basis. $\endgroup$ – Jonas Meyer May 2 '17 at 16:58

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