Linear space and cardinals Let $V$ be a linear space over a field $K$. If $L, S\subseteq V$ such that $L$ is a linear independent set with the property that $L\subseteq \mathrm{Span}(S)$, prove that: $|L|\leq |S|$, where $\leq$ is the total order defined for cardinals.
I am interested in a proof in the case in which $V$ is not finite dimensional.
Thanks!
 A: If $L\subseteq \mathrm{Span}(S)$, then $\mathrm{Span}(S)$ contains an independent set of cardinal $|L|$. 
It suffices to show that $\mathrm{Span}(S)$ cannot contain an independent set of size strictly bigger than $|S|$.
Indeed, assume it contains such a set $X$. By Zorn's Lemma, $X$ can be completed into a base $B$ of $\mathrm{Span(S)}$, because the set of sets $F$ such that $X\cup F$ is independent contains $\emptyset$, is closed under inclusion, and has an upper bound (which is $\mathrm{Span}(S)$), so it contains at least one maximal element $F_m$.
We have that $B=X\cup F_m$ is a base strictly bigger than $S$. Hence each element of $s\in S$ can be written $s=\sum_{b\in B_s} a_{s,b}b$, where $a_{s,b}\in K$ and $B_s$ is a finite subset of $B$.
Since $B$ is bigger than $S$, $B$ is also strictly bigger than $\bigcup_{s\in S} B_s$ (for this argument we need $S$ infinite). This means there is a $b\in B$ that is not used in the decompositions of elements in $S$, and therefore $S$ is independent of $b$. This is contradictory with $b\in \mathrm{Span}(S)$, so such $X$ cannot exist, and thus $|L|\leq |S|$.
