Date: Nov 14, 2012 8:57 AM
Author: Jose Carlos Santos
Subject: Re: Dimension of the space of real sequences
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