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 $$\left|{B}\right| = n$$.

Then $$0 \notin B$$, or $$B - \left\{{0}\right\}$$ would be a generator for $$G$$ with $$n-1$$ elements (as $$H$$, being linearly independent, does not contain $0$), thus contradicting 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 $$a_p = \sum_{k=1}^{p-1} \mu_k a_k$$.

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

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

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.