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 $$\bigcup_{k=1}^n S_k$$ 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 $$H \subseteq C \subseteq H \cup B$$.