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Re: Dimension of the space of real sequences
Posted:
Nov 18, 2012 12:54 PM
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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).
Regards,
David Bernier
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