
Re: Dimension of the space of real sequences
Posted:
Nov 14, 2012 12:21 AM


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 ...
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 squaresummable sequences of reals .
S is vector subspace of l^2, the real infinitedimensional 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

