
Re: Dimension of the space of real sequences
Posted:
Nov 18, 2012 12:54 PM


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).
Regards,
David Bernier

