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

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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

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:

\(\displaystyle \card {L \setminus B}\) \(\le\) \(\displaystyle \card {L \setminus L'}\) 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.


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.

$\blacksquare$


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