Size of Linearly Independent Subset is at Most Size of Finite Generator/Proof 1

Proof
We first consider the case where $L$ is finite.

Let $S \subseteq \N$ be the set of all $n \in \N$ such that:
 * For every finite generator $F$ of $V$, if $\left\vert{L \setminus F}\right\vert \le n$, then $\left\vert{L}\right\vert \le \left\vert{F}\right\vert$.

We use the Principle of Mathematical Induction to prove that $S = \N$.

Let $\left\vert{L \setminus F}\right\vert \le 0$.

Then from Cardinality of Empty Set:
 * $L \setminus F = \varnothing$

By Set Difference with Superset is Empty Set:
 * $L \subseteq F$

By Cardinality of Subset of Finite Set:
 * $\left\vert{L}\right\vert \le \left\vert{F}\right\vert$

Hence:
 * $0 \in S$

Assume the induction hypothesis that $n \in S$.

Let $F$ be a finite generator of $V$ such that:
 * $\left\vert{L \setminus F}\right\vert = n + 1$

Let $v \in L \setminus F$.

Let $L' = L \cap \left({F \cup \left\{{v}\right\}}\right)$.

By Intersection is Subset:
 * $L' \subseteq L$

By Subset of Linearly Independent Set, it follows that $L'$ is linearly independent over $R$.

Also by Intersection is Subset:
 * $L' \subseteq F \cup \left\{{v}\right\}$

Therefore, by Linearly Independent Subset of Basis of Vector Space:
 * there exists a basis $B$ of $V$ such that:
 * $L' \subseteq B \subseteq F \cup \left\{{v}\right\}$

Since $v \notin F$ is a linear combination of $F$, it follows that $F \cup \left\{{v}\right\}$ is linearly dependent over $R$.

Therefore:
 * $B \subsetneq F \cup \left\{{v}\right\}$

By Cardinality of Subset of Finite Set:
 * $\left\vert{B}\right\vert < \left\vert{F \cup \left\{{v}\right\}}\right\vert = \left\vert{F}\right\vert + 1$

Hence:
 * $\left\vert{B}\right\vert \le \left\vert{F}\right\vert$

We have that:

Since $n \in S$:
 * $\left\vert{L}\right\vert \le \left\vert{B}\right\vert \le \left\vert{F}\right\vert$

Hence:
 * $n + 1 \in S$

and so the induction step has been completed.

By Set Difference is Subset:
 * $L \setminus F \subseteq L$

From Subset of Finite Set is Finite:
 * $L \setminus F$ is finite.

Therefore, we can apply the fact that $S = \N$ to conclude that:
 * $\left\vert{L}\right\vert \le \left\vert{F}\right\vert$

Let $L$ be infinite.

Then by Set is Infinite iff exist Subsets of all Finite Cardinalities, there exists a finite subset $L' \subseteq L$ such that:
 * $\left\vert{L'}\right\vert = \left\vert{F}\right\vert + 1$

By Subset of Linearly Independent Set, it follows that $L'$ is linearly independent over $R$.

It is proven above that this is impossible.

Hence the result.