```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 SantosSuppose 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^2of square-summable sequences of reals .S is vector subspace of  l^2, the real infinite-dimensional Hilbertspace. [ 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 dimensionalreal 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.htmlUsually, when nothing is said, statements on Banach spacesapply irrespective of whether it is a real Banach spaceor 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 wantthe 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_basisBaire 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
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