Let $S$ be a subset of a vector space. I conjecture that $S$ is linearly dependent if and only if there exists a proper subset $S' \subset S$ such that $\operatorname{span}(S')=\operatorname{span}(S)$. Is this the case?
If this is the case, I see an analogy between this and the definition of infinite set as a set which is bijective with a proper subset of itself. Could someone elaborate on this if it is true?
If $S$ is linearly dependent, then we can write $\sum_k \alpha_k x_k =0$ for some $\alpha_k$ not all zero and $x_k \in S$. Without loss of generality, we can assume that $\alpha_1 \neq 0$, and so we have $x_1 \in \operatorname{sp} \{x_2,...\}$. Then let $S' = S \setminus \{x_1\}$, which is a proper subset, and clearly $\operatorname{sp} S = \operatorname{sp} S'$.
For the other direction, if $\operatorname{sp} S = \operatorname{sp} S'$ with $S'$ being a proper subset of $S$, then pick $x \in S \setminus S'$.
If $x=0$, then $S' \cup \{0\}$ is linearly dependent and hence so is $S$, otherwise we have $x =\sum_k \alpha_k x_k$ for some $x_k \in S'$. Hence $\sum_k \alpha_k x_k + (-1) x = 0$ and so $S$ is linearly dependent.