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: {{begin-eqn}} {{eqn | l = \card {L \setminus B}       | o = \le       | r = \card {L \setminus L'}       | c = Relative Complement inverts Subsets and Cardinality of Subset of Finite Set }} {{eqn | r = \card {L \setminus \paren {F \cup \set v} }       | c = Set Difference with Intersection is Difference }} {{eqn | r = \card {\paren {L \setminus F} \setminus \set v}       | c = Set Difference with Union }} {{eqn | r = n + 1 - 1       | c = Cardinality of Set Difference with Subset }} {{eqn | r = n }} {{end-eqn}} 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. {{qed}}