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Re: Dimension of the space of real sequences
Posted:
Nov 14, 2012 8:57 AM
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On 14-11-2012 10:19, quasi wrote:
>> Can someone please tell me how to prove that the real vector
>> space of all sequences of real numbers has uncountable
>> dimension?
>
> Let V be the set of infinite sequences of real numbers, regarded
> as a vector space over R.
>
> For each a in R, let v_a = (e^a,e^(2a),e^(3a),...).
>
> Let S = {v_a | a in R}.
>
> S has the same cardinality as R, hence S is uncountable.
>
> Claim the elements of S are linearly independent over R.
>
> Suppose otherwise.
>
> Let n be the least positive integer such that there exist
> n distinct real numbers a_1, ..., a_n such that
>
> v_(a_1), ..., v_(a_n)
>
> are linearly dependent over R.
>
> Without loss of generality assume a_1 < ... < a_n.
>
> Then
>
> (c_1)*(v_(a_1)) + ... + (c_n)*(v_(a_n)) = 0
>
> for some real numbers c_1, ..., c_n.
>
> By minimality of n, it follows that c_1, ..., c_n are
> all nonzero.
>
> Then
>
> (c_1)*(v_(a_1)) + ... + (c_n)*(v_(a_n)) = 0
>
> implies
>
> (c_1)*(e^(k*(a_1))) + ... + (c_n)*(e^(k*(a_n))) = 0
>
> for all k in N.
>
> Dividing both sides by e^(k*(a_n)), we get
>
> (c_1)*(e^(k*(a_1 - a_n))) + ... + c_n = 0
>
> for all k in N.
>
> Taking the limit of both sides as k -> oo yields c_n = 0,
> contradiction.
>
> Hence the elements of S are linearly independent over R, as
> claimed.
>
> It follows that the dimension of V over R is uncountable.
Thanks for the proof.
Best regards,
Jose Carlos Santos
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