I'm trying to understand this below proof.
Given $S$ a linearly independent set of vectors in $V$ and $S' \subset S$ a subset, then $S'$ is also linearly independent.
If $S$ is finite to start with or even countable, I'm fine with the proof. In the case where $S$ is uncountably infinite, I don't know if I fully buy this proof.
Suppose we have $\sum\limits_{v_{\alpha} \in S'} c_{\alpha} v_{\alpha} = 0$. By defining $c_{\alpha} = 0$ for $v_{\alpha} \in S \setminus S'$, we can interpret this as $0 = \sum\limits_{v_{\alpha} \in S} c_{\alpha} v_{\alpha}$. As $S$ is linearly independent, all the $c_{\alpha} = 0$, so $S'$ is linearly independent as well.
My problem is this. By using $v_{\alpha}$, we're "allowing" $S$ to be uncountably infinite by referencing what is in effect an arbitrary indexing set .The next step of extending the sum to $S$ really means "add $0 $ as many times as necessary." The infinite series of $0$ is convergent and sums to $0$, so that's fine, but I really don't know what adding $0$ uncountably many times looks like. I'm inclined to think there has to be some further assumptions built into $S$ or at least the set difference $S' \setminus S$. I don't want to assume that $V$ is finite-dimensional.
Best Answer
You are correct that proof is sloppy in the infinite case. Given the usual definition of linear independence of infinite sets:
it is not correct to start a proof with:
since nowhere in the definition is a sum over all elements considered. Note though, that we now essentially have two different definitions of linear independence: one for infinite sets and one for finite sets. Using this definition, a proof would go:
Suppose $S$ is linearly independent and infinite, and $S'\subseteq S$. If $S'$ is finite, $S'$ is linearly independent since $S$ is. So suppose $S'$ is infinite, and $T \subseteq S'$ is finite. Then since $S' \subseteq S$, $T$ is a finite subset of $S$, and so since $S$ is linearly independent, so is $T$. So every finite subset of $S'$ is linearly independent; so $S'$ is linearly independent as well.
Note that here you'll need a separate proof for the case where $S$ is finite, because there is a different criterion for linear independence in that case.
It's possible, though, that this proof is from a source using a different definition of linear independence and a different definition of summation, where infinite sums are allowed (assuming only finitely many terms are nonzero.) In that case, the proof would be valid.