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Re: Dimension of the space of real sequences
Posted:
Nov 14, 2012 3:33 PM
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"José 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.
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