Date: Nov 14, 2012 12:21 AM
Author: David Bernier
Subject: Re: Dimension of the space of real sequences

On 11/13/2012 05:27 PM, José Carlos Santos wrote:
> Hi all,
> Can someone please tell me how to prove that the real vector space of
> all sequences of real numbers has uncountable dimension?
> Best regards,
> Jose Carlos Santos

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 ...

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" :

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,

"(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."


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 ].

David Bernier