
Re: Dimension of the space of real sequences
Posted:
Nov 14, 2012 2:24 AM


On 14112012 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 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 ].
Thanks a lot.
Best regards,
Jose Carlos Santos

