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 message

news: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 be

zero.

> >

> > I imagine there must be lots of ways to exhibit such a set, but as a

hint

> > 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 Santos

As others have given fairly detailed examples of essentially the same

solution looking at sequences v_a = (a, a^2, a^3, ...), and since my

approach was quite different I thought I'd give more details of it - I'm not

sure my "hint" would really be enough to put someone on the track if they

weren't already thinking along my lines!

So for my approach it's easier to think of maps Q--->R rather than sequences

N--->R. Of course, Q and N have the same cardinality, so the vector spaces

are isomorphic.

Given a real r, it has an associated Dedekind cut in the rationals, and so

has an associated "characteristic function" mapping Q to R, namely:

i_r (q) = 1 if q < r

= 0 if q >= r

There are uncountably many i_r, (one for each r), and they can easily be

shown to be independent - basically if

sum[j = 1..n] (a_j * i_r_j) = 0

then just look at the j for which the real r_j is maximum, and consider the

function for rationals just below r_j. That will show that the

corresponding a_j is zero, and by repetition that all the a_j are zero...

Regards,

Mike.