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By definition, $S$ is linearly independent if for all $n>0,c_i\in F,s_i\in S$, we have that $c_1s_1 + \ldots + c_n s_n \neq 0$ whenever $\{c_i\}\neq \{0\}$.

Why do we restrict our attention to finite linear combinations? We could imagine a different definition:

Set $S$ is "infinitely linearly independent" if for any indexing set $I$ and $c_i\in F,s_i\in S$, we have that $\sum_{i \in I} c_is_i \neq 0$ whenever $\{c_i\}\neq \{0\}$.

Why isn't this definition, which allows arbitrary subsets of $S$, useful?

tba
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  • The current definition allows also arbitrary subsets of S, since some of the $c_i$s can be 0, and it is simpler. – LeeNeverGup Feb 11 '14 at 20:42
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    How about a infinite dimensional space? There you would have infinite $c$'s and a basis would consist of infinite lin independent terms. But off the bat, I wouldn't know any direct application of that... – imranfat Feb 11 '14 at 20:43
  • http://math.stackexchange.com/questions/165684/while-proving-that-every-vector-space-has-a-basis-why-are-only-finite-linear-co?rq=1 – yohBS Feb 11 '14 at 20:43
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    First of all, what does $$\sum_{i\in I} c_i s_i$$ mean when $I$ is an infinite set? – Daniel Fischer Feb 11 '14 at 20:50
  • Nice question. I also met the same problem when studying the definition of span of a set. – sleeve chen Jun 02 '16 at 19:21

1 Answers1

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In an infinite-dimensional space, you do sometimes do something very like this. But it's not quite as simple as you've put it so far. The basic problem:

How is $\sum_I c_is_i$ defined for $I$ infinite?

Well...to start out with the simplest case, how about $I=\mathbb{N}$? Then this is just summing a series, so we say $$\sum_{\mathbb{N}} c_is_i=\lim_{n\to\infty}\sum_{i=1}^n c_is_i$$ Now we see the subtleties begin to pop out: most to the point, what do we mean by that limit? Say our vector space was $\mathbb{R}$. Then the limit's just the $\varepsilon-\delta$ concept from advanced calculus; same thing for $\mathbb{R}^n$. But by remembering that over $\mathbb{R}$ many series have no sum, or have a sum that depends on the order of their terms, we realize there's another condition needed in the proposed definition of infinite linear independence: we'd better change it to "for any $I$ such that $\sum c_is_i$ converges..."

What's also come up is that I could only make sense of the infinite sum over $\mathbb{R}$, although I could over $\mathbb{Q}$ or $\mathbb{C}$ as well. So in a vector space over a general field $F$ there's no natural definition of an infinite sum: we need to decide on a notion of convergence and limits, which is essentially what it means to say we need a topology on $F$. Given such a topology, the spaces $F^n$ inherit a notion of convergence in the same way $\mathbb{R}^n$ does from that in $\mathbb{R}$.

Now this seems to be getting somewhere. Unfortunately as another answer mentioned this notion isn't of any use over a finite-dimensional vector space, since there aren't any infinite linear independent sets, so questions of infinite linear independence only really hit their stride when one begins to study infinite-dimensional vector spaces. The problem is that these spaces don't naturally inherit a topology from that on $\mathbb{R}$, instead, there are generally countless non-equivalent and reasonable choices for what "convergence" of an infinite sum means.

OK, fine, let's not get too far into that: what we've seen we need is an infinite-dimensional vector space $V$ over a field $F$ such that $F$ and $V$ both have topologies to be able to even talk about infinite sums. Let me now come from the other side with a very special case in which something like infinite linear independence is central to the study of certain vector spaces. The simplest infinite-dimensional example is the space $\ell^1$ of sequences $(x_i)_{i\in\mathbb{N}}$ of real numbers $x_i$ such that $\sum |x_i|<\infty$.

We can make sense of limits in $\ell^1$ almost exactly as we do in $\mathbb{R}$: avoiding setting too much notation, a sequence is Cauchy if eventually its differences have arbitrarily small $\sum|x_i|$. On a space like $\ell^1$ one gets a theory of bases formally very similar to that on $\mathbb{R}^n$ by requiring exactly that every vector be uniquely expressible as an infinite linear combination of basis elements. The basis here is even just like that in $\mathbb{R}^n,$ namely, the sequences $e_i$ that are $1$ in their $i$th place and $0$ elsewhere. Then it's intuitively unsurprising that the $e_i$ are infinitely linear independent and form a basis in the infinite sense. But they're nowhere close to a basis in the finite sense-a finite sum of them can never have infinitely many nonzero terms! In fact the most important infinite-dimensional vector spaces never have a countable basis in the finite-sums sense. So this notion of infinite linear independence is absolutely central for the study of infinite-dimensional vector spaces, which is a large part of the field called functional analysis.

Kevin Carlson
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  • What is this notion of linear combination/basis called? Who studied them first? – Michele De Pascalis May 19 '17 at 09:16
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    They're called Schauder bases. I have no idea who studied them first. – Kevin Carlson May 19 '17 at 17:07
  • Thanks, are you aware of any analogous concept for non-countable linear combinations? (Think Fourier Transform) – Michele De Pascalis May 19 '17 at 21:17
  • What's uncountable about the linear combinations in a Fourier transform? – Kevin Carlson May 19 '17 at 21:48
  • We can consider the Fourier Transform of a function $f(t)$ as a "distribution" of weights to be given to each imaginary exponential $e^{i\omega t}$ in order to have $f$. So $f$ could be loosely seen as a linear combination of a continuum of exponentials, one for every real pulsation $\omega$. – Michele De Pascalis May 19 '17 at 23:27
  • No, it can't. It's a linear combination, not loosely, of countably many exponentials $e^{nit}$.The non-integer periods can ben approximated by the integer ones. You wouldn't try to add up all uncountable many exponentials, since they interact with each other. The only definition of uncountable sum I've ever heard of gives an uncountable sum as the supremum of finite (or countable) subsums, which only converges in relatively boring cases. Uncountable sums in real vector spaces don't seem to interact well with the fact that the topology of the real line being determined by countable sequences. – Kevin Carlson May 20 '17 at 02:10
  • Where can I find proof/demonstration of this? – Michele De Pascalis May 20 '17 at 09:08
  • Well, one way to make my claim precise is that an uncountable sum of real numbers, defined as I just did, can only converge if all but countably many of its terms are zero, since otherwise uncountably many terms lie in one of the countably many segments $[1/(p+1),1/p]$ for some $p\in\mathbb{N}$. – Kevin Carlson May 20 '17 at 16:01