Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set

Theorem
Let $G$ be a finitely generated $K$-vector space.

Let $H$ be a linearly independent subset of $G$.

Let $F$ be a finite generator for $G$ such that $H \subseteq F$.

Then there is a basis $B$ for $G$ such that $H \subseteq B \subseteq F$.

Proof
Let $\mathbb S$ be the set of all $S \subseteq G$ such that $S$ is a generator for $G$ and that $H \subseteq S \subseteq F$.

Because $F \in \mathbb S$, it follows that $\mathbb S \ne \varnothing$.

Because $F$ is finite, then so is every member of $\mathbb S$.

Let $R = \left\{{r \in \Z: r = \left|{S}\right| \in \mathbb S}\right\}$.

That is, $R$ is the set of all the integers which are the number of elements in generators for $G$ that are subsets of $F$.

Let $n$ be the smallest element of $R$.

Let $B$ be an element of $\mathbb S$ such that $\left|{B}\right| = n$.

We note that as $H$ is a linearly independent set, it does not contain $0$ by Subset of Module Containing Identity is Linearly Dependent.

Then $0 \notin B$, or $B \setminus \left\{{0}\right\}$ would be a generator for $G$ with $n-1$ elements.

This would contradict the definition of $n$.

Let $m = \left|{H}\right|$.

Let $\left \langle {a_n} \right \rangle$ be a sequence of distinct vectors such that $H = \left\{{a_1, \ldots, a_m}\right\}$ and $B = \left\{{a_1, \ldots, a_n}\right\}$.

Suppose $B$ were linearly dependent.

By Linearly Dependent Sequence of Vector Space, there would exist $p \in \left[{2 \,.\,.\, n}\right]$ and scalars $\mu_1, \ldots, \mu_{p-1}$ such that $\displaystyle a_p = \sum_{k \mathop = 1}^{p-1} \mu_k a_k$.

As $H$ is linearly independent, $p > m$ and therefore $B\,' = B \setminus \left\{{a_p}\right\}$ would contain $H$.

Now if $\displaystyle x = \sum_{k \mathop = 1}^n \lambda_k a_k$, then:
 * $\displaystyle x = \sum_{k \mathop = 1}^{p-1} \left({\lambda_k + \lambda_p \mu_k}\right) a_k + \sum_{k \mathop = p+1}^n \lambda_k a_k$

Hence, $B\,'$ would be a generator for $G$ containing $n-1$ elements, which contradicts the definition of $n$.

Thus $B$ must be linearly independent and hence is a basis.