```Date: Nov 14, 2012 3:33 PM
Author: Mike Terry
Subject: Re: Dimension of the space of real sequences

"JosÃ© Carlos Santos" <jcsantos@fc.up.pt> wrote in messagenews:aggv5fFhrc9U1@mid.individual.net...> On 14-11-2012 0:00, Mike Terry wrote:>> >> Can someone please tell me how to prove that the real vector space of> >> all sequences of real numbers has uncountable dimension?> >> > You need to exhibit an uncountable set of vectors that are linearly> > independent - i.e. no finite linear combination of the vectors can bezero.> >> > I imagine there must be lots of ways to exhibit such a set, but as ahint> > for the approach that occured to me:  think "reals" (= "Dedekind cuts":> > uncountably many of these...) composed from rationals (countably many of> > these, like the countable number of terms in a sequence...).>> Great hint. Thanks.>> Best regards,>> Jose Carlos SantosAs others have given fairly detailed examples of essentially the samesolution looking at sequences v_a = (a, a^2, a^3, ...), and since myapproach was quite different I thought I'd give more details of it - I'm notsure my "hint" would really be enough to put someone on the track if theyweren't already thinking along my lines!So for my approach it's easier to think of maps Q--->R rather than sequencesN--->R.  Of course, Q and N have the same cardinality, so the vector spacesare isomorphic.Given a real r, it has an associated Dedekind cut in the rationals, and sohas an associated "characteristic function" mapping Q to R, namely:    i_r (q) =  1 if q < r            =  0 if q >= rThere are uncountably many i_r, (one for each r), and they can easily beshown to be independent - basically if    sum[j = 1..n] (a_j * i_r_j) = 0then just look at the j for which the real r_j is maximum, and consider thefunction for rationals just below r_j.  That will show that thecorresponding a_j is zero, and by repetition that all the a_j are zero...Regards,Mike.
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