
Re: Dimension of the space of real sequences
Posted:
Nov 14, 2012 8:57 AM


On 14112012 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

