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Let $V$ be a vector space over a field $F$ and let $S$ be a (possibly infinite) subset of $V$. Show that, if $S$ is linearly dependent, there is some proper subset $S'$ of $S$ such that $\operatorname{Span}S'=\operatorname{Span}S$.

What I had trouble with was representing $S$ (and the span of $S$) when it's infinite. What does linear dependence mean if you can't use the $v_i=a_1v_1+...+a_nv_n$ definition?

My first attempt at this problem was to let $S=\{v_1,...,v_n\}$, where $v_i=a_1v_1+...+a_{i-1}v_{i-1}+a_{i+1}v_{i+1}+...+a_nv_n$ since $S$ is linearly dependent, and $S'=\{v_1,...,v_{i-1},v_{i+1},...,v_n\}$. But of course, this doesn't work when $S$ is infinite.

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    $\begingroup$ Try $\displaystyle \sum_{v \in S} a_v v$ where $a_v \neq 0$ for only finitely many $v$. $\endgroup$ – InterstellarProbe May 15 '18 at 20:47
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Assume $S$ is linearly dependent, i.e., by definition there exist coefficients $c_v$, $v\in S$, such that almost all $c_v$ are $=0$, but not all $c_v$ are $=0$, and $\sum_{v\in S}c_vv=0$. Pick $w\in S$ with $c_w\ne 0$ and show that $S':=S\setminus \{w\}$ spans the same subspace as $S$ does.

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By definition, linear dependence means some finite linear combination gives $0$, i.e., there is a finite subset $T = \{v_1,\dotsc,v_n\} \subset S$ such that $\sum_{i=1}^n a_i v_i = 0$. With this in mind your proof idea should work.

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