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Consider a Hilbert space which is infinite dimensional. If it is separable, it is well known that an orthonormal basis will be countable, while a hamel basis will be uncountable (since it is a complete space), therefore the two bases cannot coincide. But how can we prove that those two bases cannot be equal if our space is not separable?

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  • $\begingroup$ Nothing changes: if $\{ e_{\alpha}\}$ is an ONB, then the expansion coefficients of an $x\in H$ are the scalar products, so an infinite linear combination is not at the same time a finite linear combination. $\endgroup$ – user138530 Dec 7 '14 at 1:31
  • $\begingroup$ Basically we get that any x is perpendicular to all but finitely many e_i's. However, Tomasz's answer made it clear. It was a very simple question but I didn't try to think of such a simple approach. Thank you for your answers! $\endgroup$ – Nick Kolliopoulos Dec 7 '14 at 2:04
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The argument can be made rather simpler: if you have a given ON basis $e_i$, $i\in I$, the vector $v=\sum_{n\in {\bf N}} 2^{-n}e_{i_n}$, where $i_n$ are distinct, is clearly not a finite linear combination of the $e_i$ (because if we subtract from $v$ some finite linear combination of $e_i$, we can still find some $i_n$ such that the result is not orthogonal to $e_{i_n}$).

You don't need $I$ to be countable.

On the other hand, if you have a Hilbert space of Hilbert dimension at least $\mathfrak c$, then the Hilbert dimension and Hamel dimension do coincide (but with different bases). They also coincide with the cardinality of the space, in this case.

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  • $\begingroup$ Why statement in last paragraph true i.e. if you have a Hilbert space of Hilbert dimension at least c, then the Hilbert dimension and Hamel dimension do coincide $\endgroup$ – Sushil Jun 20 '15 at 4:13
  • $\begingroup$ @Sushil: I see now that I have been a little rash. You can express every element of the space, uniquely, as an infinite linear combination of a countable set of the basis vectors. This implies that if the Hilbert dimension is $\kappa$ where $\kappa^\omega=\kappa\neq 0$ (which is true for $\kappa=\mathfrak c$, or any $\kappa=\lambda^\omega$), then the cardinality of the space is equal to $\kappa$, and therefore so is the Hamel dimension. On the other hand, the cardinality of the Hilbert space of Hilbert dimension $\aleph_\omega$ is $\aleph_\omega^\omega>\aleph_\omega$. $\endgroup$ – tomasz Jun 20 '15 at 19:23
  • $\begingroup$ @Sushil: Well, that is what I think I had in mind, anyway, considering I wrote it half a year ago. $\endgroup$ – tomasz Jun 20 '15 at 19:24
  • $\begingroup$ @Sushil: So the last statement is false: it is easy to see that the cardinality of an infinite dimensional vector space is the dimension times the cardinality of a field. So the Hilbert space of (Hilbert) dimension $\kappa$, where $\kappa>\mathfrak c$ has cofinality $\omega$ will have Hamel dimension $\kappa^\omega>\kappa$. Still, this answer is not very popular, so I'd rather not needlessly bump it just to fix this mistake. $\endgroup$ – tomasz Jun 20 '15 at 20:04
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    $\begingroup$ @Sushil: More or less, yes. $\endgroup$ – tomasz Jun 25 '15 at 15:26

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