Generator of Vector Space Contains Basis

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

Every generator for $G$:
 * $(1): \quad$ has at least $n$ elements
 * $(2): \quad$ contains a basis for $G$
 * $(3): \quad$ is a basis for $G$ iff it contains exactly $n$ elements.

Proof
From: and all we need to do is show that every infinite generator $S$ for $G$ contains a finite generator.
 * Linearly Independent Subset of Basis of Vector Space
 * Bases of Finitely Generated Vector Space
 * Basis of Vector Space is Linearly Independent and a Generator

Let $\left \langle {a_k} \right \rangle_{1 \mathop \le k \mathop \le n}$ be an ordered basis of $G$.

For each $k \in \left[{1 \,.\,.\, n}\right]$ there is a finite subset $S_k$ of $S$ such that $a_k$ is a linear combination of $S_k$.

Hence $\displaystyle \bigcup_{k \mathop = 1}^n S_k$ is a finite subset of $S$ generating $G$, for the subspace it generates contains $\left\{{a_1, \ldots, a_n}\right\}$ and hence is $G$.