```Date: Nov 19, 2012 10:22 AM
Author: David C. Ullrich
Subject: Re: Dimension of the space of real sequences

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>> <903909e2-4673-4c2c-be09-e1be2da87102@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 re-learning).??? It's trivial from the definition that the set {Chi_(0,t) : t in(0,2pi)} is linearly independent: Assume some (finite) linearcombination equals 0 and show all the coefficients vanish(draw a picture of that linear combination to make thisclear).Now any independent set can be extended to an(algebraic) basis, hence the dimension of the spacein question is at least c.>>Regards,>>David Bernier
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