Vector Space has Basis between Linearly Independent Set and Spanning Set

Theorem
Let $V$ be a vector space over a field $F$.

Let $L$ be a linearly independent subset of $V$.

Let $S$ be a set that spans $V$.

Suppose that $L \subseteq S$.

Then $V$ has a basis $B$ such that $L \subseteq B \subseteq S$.

Proof
Let $\mathscr I$ be the set of linearly independent subsets of $S$ that contain $L$, ordered by inclusion.

Note that $L \in \mathscr I$, so $\mathscr I \ne \varnothing$.

Let $\mathscr C$ be a nest in $\mathscr I$.

Let $C = \bigcup \mathscr C$.

Aiming for a contradiction, suppose that $C$ is linearly dependent.

Then there exist $v_1, v_2, \ldots, v_n \in C$ and $r_1, r_2, \ldots, r_n \in F$ such that $r_1 \ne 0$:


 * $\displaystyle \sum_{k \mathop = 1}^n r_k v_k = 0$

Then there are $C_1, C_2, \ldots, C_n \in \mathscr C$ such that $v_k \in C_k$ for each $k \in \left\{{1, 2, \ldots, n}\right\}$.

Since $\mathscr C$ is a nest, $C_1 \cup C_2 \cup \cdots \cup C_n$ must equal $C_k$ for some $k \in \left\{{1, 2, \ldots, n}\right\}$.

But then $C_k \in \mathscr C$ and $C_k$ is linearly dependent, a contradiction.

Thus $C$ is linearly independent.

By Zorn's Lemma, $\mathscr I$ has a maximal element $M$ (one that is not contained in any other element).

Since $M \in \mathscr I$, $M$ is linearly independent.

All that remains is to show that $M$ spans $V$.

Suppose, to the contrary, that there exists a $v \in V \setminus \operatorname {span} \left({M}\right)$.

Then, since $S$ spans $V$, there must be an element $s$ of $S$ such that $s \notin \operatorname {span} \left({M}\right)$.

Then $M \cup \left\{{s}\right\}$ is linearly independent.

Thus $M \cup \left\{{s}\right\} \supsetneq M$, contradicting the maximality of $M$.

Thus $M$ is a linearly independent subset of $V$ that spans $V$.

Therefore, by definition, $M$ is a basis for $V$.

Also see

 * Existence of Vector Space Bases implies Axiom of Choice