If $x\in{\rm span}(S\cup\{y\})$ and $x\notin{\rm span}(S)$, then $y\in{\rm span}(S\cup\{x\})$
This statement is simple enough if ${\rm span}$ is defined in terms of finite linear combinations: if $x=\sum_{n=1}^ka_ns_n+by$ then $b=0$ implies $x=\sum_{n=1}^ka_ns_n$ and thus $x\in{\rm span}(S)$, contrary to hypothesis, so $b\ne0$ and $y=b^{-1}(x-\sum_{n=1}^ka_ns_n)$.
However, I am working off the definition ${\rm span}(S)=\bigcap\{T\in{\scr S}:S\subseteq T\}$ (or alternatively via the "axioms"
- ${\rm span}(S)\in{\scr S}$
- $S\subseteq T\to{\rm span}(S)\subseteq {\rm span}(T)$
- $S\in{\scr S}\to{\rm span}(S)=S$),
and I am also trying to extend this theorem to left modules if possible, so that the division in the previous paragraph is not applicable. This seems like a really easy question, but I can't see how to get to the answer. If this theorem is false in modules, is there a suitable definition of "linearly independent" that is more symmetric? (My definition is that a set $S$ is linearly independent if $x\notin{\rm span}(S\setminus x)$ for each $x\in S$.)
The statement is false for arbitrary modules. To take a simple example, let the module be $\Bbb Z$ as a $\Bbb Z$-module with the usual operations, and let $S=\emptyset$, $x=2$, $y=1$. Then $x\in{\rm span}(y)=\Bbb Z$ but $y\notin{\rm span}(x)=2\Bbb Z$. The statement $x_i\notin{\rm span}(\{x_j:j\in S\setminus i\})$ corresponds to $x_i\ne\sum_{j\in S}a_jx_j$ for any finitely supported family $\{a_j:j\in S\}$, which does not match the usual symmetric notion of linear independence, $\sum_{j\in S}a_jx_j=0$ only if all $a_j=0$, in an arbitrary module (since a nontrivial representation of $0$ does not necessarily allow any one of the vectors to be written in terms of the others). To use the $\Bbb Z$ example again, $\{2,3\}$ is linearly dependent in the usual sense, because $2\cdot 3-3\cdot 2=0$, but $2\notin{\rm span}(3)$ and $3\notin{\rm span}(2)$. There is no simple interpretation of this notion of linear independence in terms of spans.
For a module over a skew field $F$ or a vector space, the statement is true, and the proof is similar to the given one. Let $T=\{z:\exists b\in F,\,z-by\in{\rm span}(S)\}$. Then it is easy to show that $T$ is a subspace, and of course it contains $y$ and the vectors in $S$. Thus ${\rm span}(S\cup\{y\})\subseteq T$, so $x\in T$ and there is a $b$ such that $x-by\in{\rm span}(S)$. If $b=0$, then $x\in{\rm span}(S)$, so $b\ne0$ and $b^{-1}$ exists. Then $y=b^{-1}(x-(x-by))$, and since $x,x-by\in{\rm span}(S\cup\{x\})$, we have $y\in{\rm span}(S\cup\{x\})$.