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I have been reading Douglas's book "Banach Algebra Techniques in Spectral Theory," and encountered a problem that confuses me:

3.19: Show that no Hilbert space has linear dimension $\aleph_0$. (Hint: Use the Baire category theorem.)

I don't see how this could possibly be true, unless "linear dimension" means something different than "the dimension of any orthonormal basis." There are certainly plenty of Hilbert spaces with countable bases. Am I missing something, or is this perhaps a typo?

PS: If the statement is, "no Hilbert space can have a countable set whose span is the full Hilbert space (without taking closures)" then that's fairly obvious: Let $\{e_{j}\}$ be such a set, we have $\mathcal{H} = \cup_{j=1}^{\infty} \text{span}\{e_1, \dots, e_j\},$ so by Baire's theorem at least one $\text{span}\{e_1, \dots, e_j\}$ has nonempty interior, but this is untrue because any ball around a vector in that span includes some vectors with nonzero $e_{j+1}$ component.

Arturo Magidin
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user_35
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    "linear dimension" means "the cardinality of a Hamel basis", that is, of a regular basis as a vector space. Not "maximal orthonormal set", which is what a Hilbert basis is. You want a set that is linearly independent, and spans (that is, every vector is a (finite) linear combination of elements of the basis). – Arturo Magidin Sep 22 '22 at 15:39
  • Yes, it means that Hilbert spaces (more generally Banach spaces) can not have a countable Hamel basis – Evangelopoulos Foivos Sep 22 '22 at 15:40

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