```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
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