## Every linearly independent subset of a vector space is contained in a basis

Let $F$ be a field, let $V$ be a nonzero $F$-vector space, and let $E \subseteq V$ be a linearly independent subset. Prove that there exists a basis for $V$ over $F$ which contains $E$.

Let $S \subseteq \mathcal{P}(V)$ be the set of all linearly independent subsets of $V$ which contain $E$. Note that $S$ is partially ordered by $\subseteq$, and is nonempty since $E \in S$.

Let $C \subseteq S$ be a chain; that is, a linearly ordered subset. We claim that $\bigcup C \in S$. First, certainly $E \subseteq \bigcup C$. To see that $\bigcup C$ is linearly independent, let $D = \{d_i\} \subseteq \bigcup C$ be a finite subset and suppose $\sum a_id_i = 0$. Since $D$ is finite and $C$ is linearly ordered, there exists an element $C_t \in C$ such that $D \subseteq C_t$. Since $C_t$ is linearly independent, $a_i = 0$ for all $i$. Thus $\bigcup C$ is linearly independent. So $\bigcup C \in S$, and in fact is an upper bound for $C$. By Zorn’s lemma, there exists a maximal element $B \in S$.

We claim that $V = \mathsf{span}\ B$. To see this, suppose there exists an element $v \in V \setminus \mathsf{span}\ B$. Suppose we have $r,s_i \in F$ and $b_i \in B$ (finitely many) such that $rv + \sum s_ib_i = 0$. If $r \neq 0$, then $v = \sum -s_ir^{-1}b_i \in \mathsf{span}\ B$, a contradiction. Thus $r = 0$, and we have $\sum s_ib_i = 0$. Since $B$ is linearly independent, $s_i = 0$ for all $i$. Thus $B \cup \{v\}$ is linearly independent and contains $E$, violating the maximalness of $B$ in $S$. Thus we have $V = \mathsf{span}\ B$.

So $B$ is a linearly independent generating set (i.e. basis) for $V$ which contains $E$.