quasi
Posts:
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Registered:
7/15/05


Re: Dimension of the space of real sequences
Posted:
Nov 14, 2012 5:19 AM


José 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, 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.
quasi

