Size of Linearly Independent Subset is at Most Size of Finite Generator/Proof 2
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
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 is Linearly Independent 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:
- $B \subsetneq F'$
By Cardinality of Subset of Finite Set and Cardinality is Additive Function:
- $\card B < \card {F'} = \card F + 1$
Hence:
- $\card B \le \card F$
We have that:
\(\ds \card {B \cap L}\) | \(\ge\) | \(\ds \card {\paren {F' \cap L} \cap L}\) | Set Intersection Preserves Subsets and Cardinality of Subset of Finite Set | |||||||||||
\(\ds \) | \(=\) | \(\ds \card {F' \cap L}\) | Intersection is Associative and Set Intersection is Idempotent | |||||||||||
\(\ds \) | \(=\) | \(\ds \card {\paren {F \cap L} \cup \paren {\set v \cap L} }\) | Intersection Distributes over Union | |||||||||||
\(\ds \) | \(=\) | \(\ds \card {\paren {F \cap L} \cup \set v}\) | Intersection with Subset is Subset | |||||||||||
\(\ds \) | \(\ge\) | \(\ds N\) | 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.
$\blacksquare$