Given a set $S$ and a field $F$ we can construct the $F$-free vector space over $S$ in the following way.

Consider the set of formal sums $$FS:=\left\{\sum\_{s\in S} \alpha\_s s\,:\, \alpha\_s=0\, \text{except for a finite number of}\, s \in S\right\}.$$
The structure of an $F$-vector space is given to $FS$ by using addition and multiplication in $F$, ie $$\sum\_{s \in S} \alpha\_s s+\sum\_{s \in S} \beta\_s s = \sum\_{s \in S}(\alpha\_s+\beta\_s) s,$$$$\alpha\left(\sum\_{s \in S} \alpha\_s s\right) :=\sum\_{s\in S}(\alpha \alpha\_s)s.$$

$FS$ is called free vector space over $S$.

The element of $FS$ for wich $\alpha\_s=1$ and $\alpha\_r=0$ if $r\neq s$ is identified with $s$. This identification embeds $S$ in $FS$ and allow us to consider $S$ as a set of generators for $FS$.

In fact, by definition, every element of $FS$ can be written as a linear combination of element of $S$. My question is the following: **how can I prove that $S$ is a basis? I mean, how can I prove linear independence?** 

I think we have to add the following condition on $FS$: given $a=\sum\_{s \in S} \alpha\_s s, b=\sum\_{s \in S} \beta\_s s$ in $FS$ then $$a=b\,\text{iff}\, \alpha\_s=\beta\_s \, \text{for all}\, s \in S.$$
(in this way, linear independence is trivial).

Is this condition necessary or not to prove linear independence for $S$?

Thanks a lot in advance.