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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.

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    $\begingroup$ Notice that a linear combination is only defined for a finite sum. $\endgroup$
    – William
    Feb 6 at 2:29
  • $\begingroup$ 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 $S$. $\endgroup$
    – Karl
    Feb 6 at 2:38
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    $\begingroup$ Try proving this by contradiction. Assume some finite subset is is linearly dependent and show this implies the entire set is linearly dependent. $\endgroup$ Feb 6 at 2:40
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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.

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