A Vector Space is Infinite-Dimensional iff its Linearly Independent Set Size is Unbounded
# Statement
Let $V$ be a Vector Space. Then $V$ is an Infinite-Dimensional Vector Space If and Only If for any $n \in \mathbb{N}$, there exists some $S \subset V$ so that $|S| > n$.
# Proof
We prove the Contraposition instead:
$(\Rightarrow)$ Suppoes $\exists n \in \mathbb{N}$ so that for all Linearly Independent $S \subset V$, we have $|S| \leq n$. We know such a Linearly Independent Set exists since The Empty Set is Linearly Independent. Now pick any such $S$. Run the process of Growing a Linearly Independent Set over $V$. This process must terminate in at most $n$ steps. This gives us a Maximal Linearly Independent Set. By A Set is a Basis iff it is a Maximal Linearly Independent Set, we have exhibited a finite Vector Space Basis for $V$, and $V$ is a Finite-Dimensional Vector Space. $\checkmark$
($\Leftarrow$) Suppose $V$ is a Finite-Dimensional Vector Space. Then there exists a Vector Space Basis $S \subset V$ so that $S$ is a Finite Set. Since Spanning Set Size bound Linearly Independent Set Size, we have that for all Linearly Independent $R \subset V$, $|R| \leq |S| =: n \in \mathbb{N}$. $\checkmark$ $\blacksquare$
TODO - do by Contraposition