- #26

vanesch

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Up to here I follow. However, you probably agree that this closure adds (a LOT) of elements to the set, ....George Jones said:The [itex]T[/itex] that I defined is an inner product space that needs to be completed (made "closed" with respect to Cauchy sequences) in order to obtain a Hilbert space.

Ah, that's perhaps the key: you have at least a DENSE, countable subset: namely your basis vectors, say, with rational coefficients.

I vaguely remember that there is a theorem that says that a hilbert space that has a countable dense subset is separable, is that the point ?

cheers,

Patrick.

EDIT: I'm beginning to see the point. The basis of the closure is NOT the closure of the basis (which is uncountable)

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