How to prove the linear indenpendency of an infinite set

[julypraise]

First let me give you some definitions for the clarification of the problem:

Definition (Roman, Linear Algebra)
A nonempty set S of a vector space V with a field F is linearly independent if for any distinct vectors $s_{1} , \dots , s_{n}$ for all $a_{1} , \dots , a_{n}$ if $a_{1}u_{1} + \cdots + a_{n}u{n} = 0$ then $a_{1} = \cdots = a_{n} = 0$.

But by this definition, I don't think I can handle an infinite set S (but you may!). When S is infinite, the n up there should be infinite too. Then everything just becomes confusing to me. Can n actually be inifinite? Isn't it an error in the sense of the first order theory? Shouldn't the definition be modifed somewhat to be more rigorous, and to be fitter to deal with the set S in the case where it is infinite? Shouldn't there be any problem of countability of n when S is infinite?

I can't even properly state what actually I don't know. Please give me your helpful comment. Thanks.

 

[Fredrik]

n is finite even when S isn't. The definition of "linearly independent" is better stated like this: A set $S\subset V$ is said to be linearly independent if for all $n\in\mathbb N$, all $s_1,\dots,s_n\in S$ and all $a_1,\dots,a_n\in F$, $$a_1 s_1+\cdots+a_n s_n=0\ \Rightarrow\ a_1=\cdots=a_n=0.$$

 

[micromass]

If one knows linear independence for finite sets, then you can easily extent it to infinite sets:

An infinite set is linear independent if all its finite subsets are linear independent.

 

[Deveno]

one can simply say a set (infinite OR finite) is linearly independent if any finite subset is linearly independent. when the set in question is finite, this, of course, means we must test the entire set for linear independence.

(why just finite subsets? well this has to do with the fact that in a general vector space, infinite sums are undefined. however, if we have "extra structure" on a vector space (such as a topology), then we can define (for example) a notion of convergence of infinite sums, in which case it becomes meaningful to test infinite sets for linear independence. and such things do exist).

 

[blue_raver22]

I would approach this problem by first acknowledging that the definition of linear independence given above is for a finite set of vectors. However, it can be extended to infinite sets by using the concept of convergence.

To prove the linear independence of an infinite set S, we can use the following steps:

1. Assume that S is linearly dependent. This means that there exists a finite subset of S, say {s_1, s_2, ..., s_n}, such that a_1s_1 + a_2s_2 + ... + a_ns_n = 0, where at least one of the coefficients a_i is non-zero.

2. Since S is infinite, there must be an infinite number of vectors in S that are not in the finite subset {s_1, s_2, ..., s_n}. Let's call this subset T.

3. Now, consider the sequence of vectors in T: s_{n+1}, s_{n+2}, ..., s_{n+k}, ... As n increases, the number of vectors in this sequence also increases.

4. Since the set T is infinite, it is possible to find a subsequence of this sequence that converges to some vector in S. Let's call this vector s.

5. Since s_{n+k} converges to s, we can write s_{n+k} as s + e_k, where e_k is a small error term. This means that as k approaches infinity, e_k approaches 0.

6. Now, we can rewrite the linear dependence relation from step 1 as: a_1s_1 + a_2s_2 + ... + a_ns_n + a_{n+1}(s + e_{n+1}) + ... + a_{n+k}(s + e_{n+k}) + ... = 0.

7. As k approaches infinity, the error terms e_{n+1}, ..., e_{n+k}, ... approach 0. This means that the above equation reduces to a_1s_1 + a_2s_2 + ... + a_ns_n + a_{n+1}s + ... = 0.

8. But we had assumed that the coefficients a_i are non-zero, which means that the vectors s_1, s_2, ..., s_n, s are linearly dependent. This contradicts our assumption that S is linearly dependent.