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 $\card {L \setminus F} \le n$, then $\card L \le \card F$

where:
 * $L \setminus F$ denotes the set difference between $L$ and $F$
 * $\card L$ and $\card F$ denote the cardinality of $L$ and $F$ respectively.

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

Basis of the Induction
Let $\card {L \setminus F} \le 0$.

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

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

By Cardinality of Subset of Finite Set:
 * $\card L \le \card F$

Hence:
 * $0 \in S$

This is the basis for the induction.

Induction Hypothesis
It is to be shown that if $k \in S$ where $k \ge 1$, then it follows that $k + 1 \in S$.

This is the induction hypothesis:
 * For every finite generator $F$ of $V$, if $\card {L \setminus F} \le k$, then $\card L \le \card F$

It is to be demonstrated that it follows that:
 * For every finite generator $F$ of $V$, if $\card {L \setminus F} \le k + 1$, then $\card L \le \card F$

Induction Step
This is the induction step:

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

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

Let $v \in L \setminus F$.

Let $L' = L \cap \paren {F \cup \set v}$.

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:
 * $L' \subseteq F \cup \set v$

Therefore, by Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set:
 * there exists a basis $B$ of $V$ such that:
 * $L' \subseteq B \subseteq F \cup \set v$

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

Therefore:
 * $B \subsetneq F \cup \set v$

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

Hence:
 * $\card B \le \card F$

We have that:

Since $n \in S$:
 * $\card L \le \card B \le \card F$

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:
 * $\card L \le \card F$

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:
 * $\card {L'} = \card F + 1$

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.