# Size of Linearly Independent Subset is at Most Size of Finite Generator

## 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:

$\size L \le \size F$

## Proof 1

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:

$L \setminus F = \O$
$L \subseteq F$
$\card L \le \card F$

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

$L' \subseteq L$

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

Also by Intersection is Subset:

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

$B \subsetneq F \cup \set v$
$\card B < \card {F \cup \set v} = \card F + 1$

Hence:

$\card B \le \card F$

We have that:

 $\displaystyle \card {L \setminus B}$ $\le$ $\displaystyle \card {L \setminus L'}$ Relative Complement inverts Subsets and Cardinality of Subset of Finite Set $\displaystyle$ $=$ $\displaystyle \card {L \setminus \paren {F \cup \set v} }$ Set Difference with Intersection is Difference $\displaystyle$ $=$ $\displaystyle \card {\paren {L \setminus F} \setminus \set v}$ Set Difference with Union $\displaystyle$ $=$ $\displaystyle n$ (?)

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.

$L \setminus F \subseteq L$
$L \setminus F$ 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.

$\blacksquare$

## Proof 2

Let $S \subseteq \N$ be the set of all natural numbers $n \in \N$ such that:

For any finite generator $F$ of $V$ over $R$, if $\card {F \cap L} \ge n$, then $\card L \le \card F$.

It is to be demonstrated that $S = \N$.

That is, that $\card {F \cap L} \ge n \implies \card L \le \card F$ for all $n \in \N$.

By Intersection is Subset and Cardinality of Subset of Finite Set, we have $\card {F \cap L} \le \card F$.

Hence, it is vacuously true that $\card F + 1 \in S$.

Therefore, $S$ is non-empty.

From the well-ordering principle, $S$ has a smallest element $N$.

If $N = 0$, the theorem immediately follows.

Aiming for a contradiction, suppose $N \ge 1$.

Let $\card {F \cap L} \ge N - 1$.

If $L \subseteq F$, the theorem follows from Cardinality of Subset of Finite Set.

Otherwise, there exists a $v \in L$ such that $v \notin F$.

Let $F' = F \cup \set v$.

By Intersection is Subset, we have $F' \cap L \subseteq L$, so it follows by Subset of Linearly Independent Set that $F' \cap L$ is linearly independent over $R$.

Also, by Intersection is Subset:

$F' \cap L \subseteq F'$

We have that $F'$ is a generator of $V$ over $R$.

By Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set, there exists a basis $B$ for $V$ such that:

$F' \cap L \subseteq B \subseteq F'$

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

By the definition of a basis:

$B \subsetneq F'$
$\card B < \card {F'} = \card F + 1$

Hence:

$\card B \le \card F$

We have that:

 $\displaystyle \card {B \cap L}$ $\ge$ $\displaystyle \card {\paren {F' \cap L} \cap L}$ Set Intersection Preserves Subsets and Cardinality of Subset of Finite Set $\displaystyle$ $=$ $\displaystyle \card {F' \cap L}$ Intersection is Associative and Intersection is Idempotent $\displaystyle$ $=$ $\displaystyle \card {\paren {F \cap L} \cup \paren {\set v \cap L} }$ Intersection Distributes over Union $\displaystyle$ $=$ $\displaystyle \card {\paren {F \cap L} \cup \set v}$ Intersection with Subset is Subset $\displaystyle$ $\ge$ $\displaystyle N$ Cardinality is Additive Function

Since $N \in S$, it follows by the definition of a basis that:

$\card L \le \card B \le \card F$

Hence:

$N - 1 \in S$

But this contradicts the assumption that $N$ is the smallest element of $S$.

Therefore:

$N = 0$

Hence the result.

$\blacksquare$

## Proof 3

Let $\alpha_1, \alpha_2, \ldots, \alpha_n$ be a generator of $V$.

Let $\xi_1, \xi_2, \ldots, \xi_r$ be a linearly independent set of elements of $V$.

Hence the sequence $\sequence {\xi_1, \alpha_1, \alpha_2, \ldots, \alpha_n}$ is a linearly dependent sequence of elements of $V$.

One of these elements, which cannot be $\xi_1$, is a linear combination of the preceding elements.

Let this elements be $\alpha_i$.

So we can omit $\alpha_i$ from that sequence, and the remaining set is still a generator of $V$.

Therefore $\xi_2$ is a linear combination of these.

Thus $\sequence {\xi_2, \xi_1, \alpha_1, \alpha_2, \ldots, \alpha_{i - 1}, \alpha {i + 1}, \ldots, \alpha_n}$ is a linearly dependent sequence of elements of $V$.

Again, one of them is a linear combination of the preceding elements.

This cannot be $\xi_2$, as $\xi_2$ has no preceding elements.

Neither can it be $\xi_1$, as $\xi_1$ and $\xi_2$ are linearly independent.

Thus we can omit whichever $\alpha_j$ it is, and we have a new set which is a generator of $V$.

This consists of $\xi_1$, $\xi_2$ and whichever $n - 2$ of the remaining elements of $\set {\alpha_1, \alpha_2, \ldots, \alpha_n}$.

After $p$ such steps, we have a set which is a generator of $V$ which consists of:

$\xi_1, \xi_2, \ldots, \xi_p$ and $n - p$ of the elements of $\set {\alpha_1, \alpha_2, \ldots, \alpha_n}$.

Aiming for a contradiction, suppose suppose $n < r$.

Then when $p = n$, the remaining set which is a generator of $V$ consists of:

$\xi_1, \xi_2, \ldots, \xi_n$

and there is at least one more element $\xi_{n + 1}$.

This is a linear combination of $\set {\xi_1, \xi_2, \ldots, \xi_n}$.

But this contradicts the supposition that $\set {\xi_1, \xi_2, \ldots, \xi_n, \xi_{n + 1} }$ is a linearly independent set.

Hence, by Proof by Contradiction, $n \ge r$.

The result follows.

$\blacksquare$