```Date: Nov 14, 2012 5:19 AM
Author: quasi
Subject: Re: Dimension of the space of real sequences

JosÃ© Carlos Santos 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, regardedas 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)) = 0for some real numbers c_1, ..., c_n.By minimality of n, it follows that c_1, ..., c_n areall nonzero.Then   (c_1)*(v_(a_1)) + ... + (c_n)*(v_(a_n)) = 0implies   (c_1)*(e^(k*(a_1))) + ... + (c_n)*(e^(k*(a_n))) = 0for 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.quasi
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