Linearly Independent Set is Contained in some Basis

Theorem
Let $G$ be a vector space of $n$ dimensions.

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

There exists a basis $B$ for $G$ such that $H \subseteq B$.

Proof
By hypothesis there is a basis $B$ of $G$ with $n$ elements.

Then $H \cup B$ is a generator for $G$.

So by Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set there exists a basis $C$ of $G$ such that $H \subseteq C \subseteq H \cup B$.