Linearly Independent Set Size is Unbounded iff there exists an Infinite Linearly Independent Set
# Statement
Let $V$ be a Vector Space. Then $\forall n \in \mathbb{N}$ there exists Linearly Independent $S \subset V$ so that $|S| > n$ If and Only If there exists Linearly Independent $R \subset V$ so that $|R| = \infty$.
# Proof
$(\Leftarrow)$ $R$ satisfies the requirement for any $n \in \mathbb{N}$. $\checkmark$
$(\Rightarrow)$ This means $V$ is an Infinite-Dimensional Vector Space because A Vector Space is Infinite-Dimensional iff its Linearly Independent Set Size is Unbounded. Therefore, starting from $\emptyset$, the Growing a Linearly Independent Set process will not terminate, otherwise by A Set is a Basis iff it is a Maximal Linearly Independent Set, we would exhibit a finite Vector Space Basis and $V$ would be a Finite-Dimensional Vector Space. Now consider the Set $$R = \bigcup\limits_{n \in \mathbb{N}} {S_{n}}$$ where $S_{n}$ is the Linearly Independent set produced by the Growing a Linearly Independent Set process at step $n \in \mathbb{N}$. Since the process does not terminate and $|S_{n}| = n$, we have that $|R| = \infty$. We claim that $R$ is Linearly Independent.
Suppose $R$ is Linearly Dependent. Then there is some $m \in \mathbb{N}$, nontrivial $c_{1}, \dots, c_{m} \in F$, distinct $\mathbf{a}{1}, \dots, \mathbf{a}{m} \in R$ so that $$c_{1} \mathbf{a}{1} + \cdots + c{m} \mathbf{a}{m} = \mathbf{0}.$$ But there must be some $N \in \mathbb{N}$ so that all $\mathbf{a}{1}, \dots, \mathbf{a}{m} \in S{N}$. Therefore $S_{N}$ is Linearly Dependent, contradicting our assumption that $S_{n}$ are Linearly Independent for all $n \in \mathbb{N}$. Therefore, $R$ is an infinite Linearly Independent Set. $\checkmark$ $\blacksquare$