I need help in the proof of the following theorem:
If $V = \text{span}(S)$, $S$ a nonempty subset of $V$, then there is a basis $S_{0}$ of $V$ such that $S_{0} \subseteq{S}$.
Here's my attempt to solve:
$\text{Proof}:$ Let $\Sigma = \{ T : T \subseteq S, \text{T is a L.I. subset of V}$} ordered by inclusion, i.e, $ A \leq B \iff A \subseteq B$. Let $\{S_{i}\}_{i \in I}$ an arbitrary chain in $\Sigma$. Then, defining $\tilde{S} = \bigcup_{i \in I}S_{i}$, we have that $S_{i} \subseteq \tilde{S}$, $\forall i \in I$. Therefore $\tilde{S}$ is a upper bound. Thus, follows by Zorn's Lemma that $\exists S_{0} \in \Sigma$ such that $S_{0}$ is maximal. Since $S_{0} \subseteq S$ is maximal and LI, follows that $S_{0}$ is a basis for $V$.
Is this wrong? What is missing?
Thanks in advance.