A Vector Space is Finite-Dimensional iff it has a Finite Spanning Set
# Statement
Let $V$ be a Vector Space. Then there is some finite $S \subset V$ for which $\text{span} S = V$ If and Only If $V$ is a Finite-Dimensional Vector Space.
# Proof 1
$(\Rightarrow)$ Consider the following construction
- Start with $R = \emptyset$.
- At each step, add to $R$ an element from $S$ that is not in $\text{span} R$
At the beginning $R = \emptyset$ and the Empty Set trivially is Linearly Independent. By Growing a Linearly Independent Set, if $R$ is Linearly Independent, then each successive step will preserve that. This process must terminate since $|S| < \infty$. When it terminates, we will have $S \subset \text{span} R$, so $\text{span} S \subset \text{span} R$ since The Span of subset of a Span is also a subset of that Span. But $$V = \text{span} S \subset \text{span} R \subset V,$$ so we have $\text{span} R = V$ and $R$ is a Vector Space Basis for $V$. Since $R \subset S$, we have that $|R| \leq |S| < \infty$, so it is a Finite Set. Therefore $V$ is a Finite-Dimensional Vector Space. $\checkmark$
$(\Leftarrow)$ If $V$ is Finite-Dimensional Vector Space, then it has a finite Vector Space Basis. A Vector Space Basis is by definition a spanning set of $V$. $\checkmark$
$\blacksquare$
# Proof 2 *
$(\Rightarrow)$ If $V$ has a finite spanning set, then it has a Minimal Spanning Set, since we can remove elements one-by-one from $S$ until it is either empty or it is no longer a Spanning Set. Call this Minimal Spanning Set $R$. Because A Set is a Basis iff it is a Minimal Spanning Set, $R$ must be a Vector Space Basis for $V$. Since $R \subset S$, we have that $|R| \leq |S| < \infty$, so $V$ has a finite Vector Space Basis and is thus finite-dimensional. $\checkmark$
$(\Leftarrow)$ If $V$ is finite-dimensional, then it has a finite Vector Space Basis which is a Spanning Set by definition. $\checkmark$ $\blacksquare$