# Any finite subset of a linearly independent set is linearly independent

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.

• Notice that a linear combination is only defined for a finite sum. Feb 6 at 2:29
• Adding to William's comment: it sounds like you might not have the correct definition in mind for "linearly independent" as it applies to $S$. You only need to prove something of finite linear combinations of elements of $S$.
– Karl
Feb 6 at 2:38
• Try proving this by contradiction. Assume some finite subset is is linearly dependent and show this implies the entire set is linearly dependent. Feb 6 at 2:40

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.