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It is well known that every separable Hilbert space has a countable orthonormal basis. This type of basis is a schauder basis. If the Hilbert space is nonseperable, the Hilbert space has a orthonormal basis wich is not countable. By definition, it isn't a Hamel basis and it isn't a Schauder basis, right? What kind of basis is this? There has to be a definition of basis which doesn't assume "countability", but I only know "Hamel basis" and "Schauder basis". Could you tell me, what kind of basis a nonseperable Hilbert space has? Regards

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I'm not sure whether this helps you, it's taken from Weidmanns Book about Hilbert spaces (which is in German). He defines an orthogonal basis of a Hilbert space $H $ (without introducing a special name for it other than that) as a system of vectors $M=\{e_i\}_{i\in X},e_i\in H$ for some index set $X$ such that

  1. $\langle e_i, e_j\rangle = \delta_{ij}$
  2. $H \subset \overline{L(M)}$

where $L(M)$ is the set of finite linear combinations of $e_i$. Here $X$ is allowed to have arbitrary cardinality. Note that this allows infinite sums $\sum_{i\in X} a_i e_i $ (taking the closure makes this a non algebraic definition), but these only make sense if at most countable many $a_i$ are nonzero.

He then shows that every Hilbert space has such an orthonormal base. This does not seem to be true for pre Hilbert spaces. He also shows that $H$ is separable iff $X$ is countable and that the cardinality of each two orthogonal bases is the same.

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  • $\begingroup$ okay, thank you, your answer is helpful. I'm from Germany and I didn't know this book. I always use "Werner- Funktionalanalysis", maybe you know it:). Regards $\endgroup$ – NewBunny Feb 25 '15 at 14:14
  • $\begingroup$ @NewBunny If the anwser really is helpful then you may consider upvoting my answer, unless you consider this request as shameless. I don't really mind, but if you continue to ask questions here and forget to appreciate (good) answers you may, some day, get no more answers at all... $\endgroup$ – Thomas Feb 25 '15 at 15:01
  • $\begingroup$ hi, yes I know and I will do that. But I couldn't finish signing up / registration for my account at the moment because I forgot my passwort of my email-adress to confirm my registration. In a few days I'm back at home and then I will finish my registration and then I'm able to vote up the answer. At the moment, I can't do this. But don't worry, I will vote up all the answers here which are helpful, and your answer I will vote up too $\endgroup$ – NewBunny Feb 25 '15 at 15:15

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