# Uncountable infinite dimensional Hilbert space

vanesch
Staff Emeritus
Gold Member
George Jones said:
The $T$ 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.

Up to here I follow. However, you probably agree that this closure adds (a LOT) of elements to the set, ....
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)

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

Bravo! (is NOT --> *may be* NOT), hence the additional difficulty.

Seratend.

George Jones
Staff Emeritus
Gold Member
seratend said:
With such a deduction a real number which is a cauchy sequence of rational numbers implies |R is a countable set from your deduction.
No. And you mean an equivalence class of Cauchy sequences of rational numbers.

seratend said:
In addition, you seem to mix the separability with the closure and the existence of a basis ("This has to be done in order for a countable basis to exist") of an hilbert space that may be countable or not countable and always exists thanks the axiom of choice.
By snipping the context of the quote, you make me appear to say something that I did not. In particular, I did not say that there exists a countable orthonormal basis for the Hilbert space completion of every inner product space.

Regards,
George

Seratend.

George Jones
Staff Emeritus