Date: Nov 14, 2012 2:24 AM
Author: Jose Carlos Santos
Subject: Re: Dimension of the space of real sequences

On 14-11-2012 5:21, David Bernier wrote:

>> Can someone please tell me how to prove that the real vector space of
>> all sequences of real numbers has uncountable dimension?

>
> Suppose the n't term of some sequence is s_n.
>
> Then the set S of vectors 's' such that s_n = O(1/n^2)
> is also vector space ...
>
>
> Hint:
> s_n = O(1/n^2) means "Exists K, a real number, |s_n| <= K * (1/n^2) " .
>
> So S is a real vector space.
> All elements of S are elements of the real Hilbert space l^2
> of square-summable sequences of reals .
>
> S is vector subspace of l^2, the real infinite-dimensional Hilbert
> space. [ infinite dimensional: with Hilbert spaces, one normally
> classifies spaces by the cardinality of a Hilbert basis of
> the Hilbert space].
>
>
> By what appears below, a Hamel basis (vector space basis)
> of the infinite dimensional
> real vector space l^2 does not have a countable basis.
> So S does not have a countable basis.
>
>
> Background at PlanetMath
> =========================
>
> There's a Panet Math entry with title:
> "Banach spaces of infinite dimension do not have a countable Hamel basis" :
> http://planetmath.org/encyclopedia/ABanachSpaceOfInfiniteDimensionDoesntHaveACountableAlgebraicBasis.html
>
>
> Usually, when nothing is said, statements on Banach spaces
> apply irrespective of whether it is a real Banach space
> or a complex one ...
>
> The proof at PlanetMath appeals to the Baire Category Theorem,
> http://en.wikipedia.org/wiki/Baire_category_theorem
>
> "(BCT1) Every complete metric space is a Baire space."
>
> As I recall, when using the form I know of BCT, we always want
> the metric space to be complete.
>
>
> Banach spaces are complete normed spaces .
>
>
>
> They cite a Monthly article:
> 1 H. Elton Lacey, The Hamel Dimension of any Infinite Dimensional
> Separable Banach Space is c, Amer. Math. Mon. 80 (1973), 298.
>
> ( c = cardinality of the continuum ).
>
> A Hamel basis is a basis in the sense of vector spaces (only finite sums).
>
> A Hilbert basis for a Hilbert space :
> The Wikipedia article on Orthonormal basis says this about
> "Hilbert basis":
>
> "Note that an orthonormal basis in this sense is not generally a Hamel
> basis, since infinite linear combinations are required."
>
> cf.:
>
> http://en.wikipedia.org/wiki/Orthonormal_basis
>
>
> Baire space: [at Wikipedia] Any countable intersection of dense open
> sets, is itself dense.
>
> N.B.: There's a way of switching things around by looking
> at the complements, which are then closed sets.
> [ By De Morgan's laws, if I'm not mistaken ].


Thanks a lot.

Best regards,

Jose Carlos Santos