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, JosÃ© 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 ...

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 ].

David Bernier