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:
- $\size L \le \size F$
Proof
We first consider the case where $L$ is finite.
Let $S \subseteq \N$ be the set of all $n \in \N$ such that:
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:
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 is Linearly Independent, 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:
\(\ds \card {L \setminus B}\) | \(\le\) | \(\ds \card {L \setminus L'}\) | Relative Complement inverts Subsets and Cardinality of Subset of Finite Set | |||||||||||
\(\ds \) | \(=\) | \(\ds \card {L \setminus \paren {F \cup \set v} }\) | Set Difference with Intersection is Difference | |||||||||||
\(\ds \) | \(=\) | \(\ds \card {\paren {L \setminus F} \setminus \set v}\) | Set Difference with Union | |||||||||||
\(\ds \) | \(=\) | \(\ds n + 1 - 1\) | Cardinality of Set Difference with Subset | |||||||||||
\(\ds \) | \(=\) | \(\ds n\) |
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.
- $L \setminus F \subseteq L$
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 is Linearly Independent, it follows that $L'$ is linearly independent over $R$.
It is proven above that this is impossible.
Hence the result.
$\blacksquare$
Sources
- 1964: Iain T. Adamson: Introduction to Field Theory ... (previous) ... (next): Chapter $\text {I}$: Elementary Definitions: $\S 4$. Vector Spaces: Theorem $4.1$
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Theorem $27.9$