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

Theorem
Let $R$ be a division ring.

Let $V$ be an $R$-vector space.

Let $F \subseteq V$ be a finite generator of $V$ over $R$.

Let $L \subseteq V$ be linearly independent over $R$.

Then $\left\vert{L}\right\vert \le \left\vert{F}\right\vert$.

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 $S = \N$.

If $\left\vert{L \setminus F}\right\vert \le 0$, then it follows by Cardinality of Empty Set that $L \setminus F = \varnothing$.

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 Subset, we have $L' \subseteq L$; by Subset of Linearly Independent Set, it follows that $L'$ is linearly independent over $R$.

Also by Intersection Subset, we have $L' \subseteq F \cup \left\{{v}\right\}$.

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

We have:

Since $n \in S$, it follows that $\left\vert{L}\right\vert \le \left\vert{B}\right\vert$.

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, it follows that $\left\vert{L}\right\vert \le \left\vert{B}\right\vert < \left\vert{F \cup \left\{{v}\right\}}\right\vert = \left\vert{F}\right\vert + 1$.

Hence, $\left\vert{L}\right\vert \le \left\vert{F}\right\vert$.

That is, $n + 1 \in S$, and so the induction step has been completed.

By Set Difference is Subset, we have $L \setminus F \subseteq L$.

Since a subset of a finite set is finite, it follows that $L \setminus F$ is finite.

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

If $L$ is infinite, then by Infinite iff Subsets Can Have Any Finite Cardinality, 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.