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 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.
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