Results concerning Generators and Bases of Vector Spaces

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

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

Every linearly independent subset of $G$:
 * $(1)$: has at most $n$ elements
 * $(2)$: is contained in a basis for $G$;
 * $(3)$: is a basis for $G$ iff it contains exactly $n$ elements.

Proof

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

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


 * A linearly independent subset of $G$ has at most $n$ elements by Linearly Independent Subset of Finitely Generated Vector Space is Finite, and is itself a basis iff it has exactly $n$ elements by Bases of Finitely Generated Vector Space and Basis of Vector Space is Linearly Independent and a Generator.

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

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

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

So by Linearly Independent Subset of Basis of Vector Space there exists a basis $C$ of $G$ such that $H \subseteq C \subseteq H \cup B$.