
Re: Dimension of the space of real sequences
Posted:
Nov 19, 2012 10:22 AM


On Sun, 18 Nov 2012 12:54:01 0500, David Bernier <david250@videotron.ca> wrote:
>On 11/15/2012 05:14 PM, W^3 wrote: >> In article >> <903909e246734c2cbe09e1be2da87102@y8g2000yqy.googlegroups.com>, >> Butch Malahide<fred.galvin@gmail.com> wrote: >> >>> On Nov 15, 7:44 am, David C. Ullrich<ullr...@math.okstate.edu> wrote: >>>> On Wed, 14 Nov 2012 18:19:29 0800, W^3<82nd...@comcast.net> wrote: >>>>> If R^N had a countable basis, then so would every subspace of R^N. In >>>>> particular l^2 would have a countable basis, call it {v_1,_2, ...}. >>>>> Setting V_n = span {v_1, ..., v_n}, we then have l^2 = V_1 U V_2 U ... >>>>> But this violates Baire, as l^2 is complete (in its usual metric) and >>>>> each V_n is closed and nowhere dense in l^2. >>>> >>>> Very good. I thought there should be something more analytic or >>>> cardinalitic instead of the (very nice) algebraic trickery that's >>>> been given. >>> >>> However, it seems to me that the "algebraic trickery" shows that there >>> is no basis of cardinality less than the continuum, whereas using >>> Baire category only shows that there is no countable base. >> >> Let's do this instead: l^2 is isomorphic to L^2([0,2pi]) (as vector >> spaces and much more), and the set {Chi_(0,t) : t in (0,2pi)} is >> linearly independent in L^2([0,2pi]). > >Yes, I have been thinking. > >As a way of computing the dimension of the span of >`n' vectors from the set {Chi_(0,t) : t in (0,2pi)}: >We choose an `n, then figure out: > dim(span({w_1, ... w_n})), > {w_1, ... w_n} being distinct vectors in the set > {Chi_(0,t) : t in (0,2pi)}. > >n=1: span has dimension 1 (trivial). >n =2 : if sometimes span had dimension < 2, > then {w_1, w_2} are not linear independent sometimes. > w_2 = alpha w_1 , alpha a scalar. > But all the Chi_(0,t) vectors are almost everywhere > 0, or 1 [ we have measurable functions before modding out > by the almost everywhere 0 functions ]. > so, alpha = 1 (contradicts w_1 =/= w_2) > >n = 3: to be continued (I'm relearning).
??? It's trivial from the definition that the set {Chi_(0,t) : t in (0,2pi)} is linearly independent: Assume some (finite) linear combination equals 0 and show all the coefficients vanish (draw a picture of that linear combination to make this clear).
Now any independent set can be extended to an (algebraic) basis, hence the dimension of the space in question is at least c.
> >Regards, > >David Bernier

