Prove that if each finite subset of a set $S$ of vectors is linearly independent, then $S$ is also linearly independent.
My attempt thus far:
Suppose that each finite subset of $S$ is linearly independent. That is, for each such subset $S_n$ of $S$, with vectors $x_1,…,x_n \in S_n$, we know that
$$a_1x_1 + … + a_nx_n = 0 \implies a_1 = … a_n = 0$$
for scalars $a_1, …, a_n$.
At this point I'm stuck. I'm not sure how to proceed with this known information and reach the conclusion that $S$ itself is also linearly independent. I know that this question has been asked elsewhere (such as for example here: Prove that any finite subset of a linearly independent set is linearly independent), but I am not understanding the answers and explanations that have been provided on these other pages and would like some additional guidance on completing this proof.
Best Answer
Linear dependence/independence can be defined in countably infinite vector spaces for any family of its elements:
The family $u_i, i \in I$ is linearly dependent if there exists a non-empty finite subset $J \subset I$ and a family of $a_i$ coefficients from corresponding field, all non-zero, such that $\sum\limits_{i \in J}a_iu_i=0\quad (1)$.
Now if we assume, that each finite subset of a set $S$ of vectors is linearly independent, but $S$ itself is linearly dependent, then from above definition we obtain non-empty finite subset with $(1)$ property, which contradicts assumption.