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, 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