Leigh.Samphier/Sandbox/Matroid Base Axiom Implies Sets Have Same Cardinality

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Theorem

Let $S$ be a finite set.

Let $\mathscr B$ be a non-empty set of subsets of $S$.


Let $\mathscr B$ satisfy the base axiom:

\((\text B 1)\)   $:$     \(\displaystyle \forall B_1, B_2 \in \mathscr B:\) \(\displaystyle x \in B_1 \setminus B_2 \implies \exists y \in B_2 \setminus B_1 : \paren {B_1 \setminus \set x} \cup \set y \in \mathscr B \)             


Then:

$\forall B_1, B_2 \in \mathscr B : \card{B_1} = \card{B_2}$

where $\card{B_1}$ and $\card{B_2}$ denote the cardinality of the sets $B_1$ and $B_2$ respectively.


Proof

Aiming for a contradiction, suppose:

$\exists B_1, B_2 \in \mathscr B : \card{B_1} \ne \card{B_2}$


Without loss of generality and from Max Equals an Operand let:

$\card{B_1 \cap B_2} = \max \set{\card{C_1 \cap C_2} : C_1, C_2 \in \mathscr B : \card{C_1} \ne \card{C_2}}$


Without loss of generality let:

$\card{B_1} > \card{B_2}$


From Set Difference of Larger Set with Smaller is Not Empty:

$\exists x \in B_1 \setminus B_2$

From base axiom $(\text B 1)$:

$\exists y \in B_2 \setminus B_1 : \paren {B_1 \setminus \set x} \cup \set y \in \mathscr B$


Let $B_3 = \paren {B_1 \setminus \set x} \cup \set y$.

We have:

\(\displaystyle \card{B_3 \cap B_2}\) \(=\) \(\displaystyle \card{\paren{\paren {B_1 \setminus \set x} \cup \set y} \cap B_2}\)
\(\displaystyle \) \(=\) \(\displaystyle \card{\paren{\paren {B_1 \setminus \set x} \cap B_2} \cup \paren {\set y \cap B_2} }\) Intersection Distributes over Union
\(\displaystyle \) \(=\) \(\displaystyle \card{\paren{\paren {B_1 \setminus \set x} \cap B_2} \cup \set y }\) Intersection with Subset is Subset
\(\displaystyle \) \(=\) \(\displaystyle \card{\paren{\paren {B_1 \cap B_2 } \setminus \set x } \cup \set y }\) Intersection with Set Difference is Set Difference with Intersection
\(\displaystyle \) \(=\) \(\displaystyle \card{\paren {B_1 \cap B_2 } \cup \set y }\) Set Difference with Disjoint Set
\(\displaystyle \) \(>\) \(\displaystyle \card{B_1 \cap B_2 }\) Cardinality of Proper Subset of Finite Set


This contradicts the choice of $B_1$ and $B_2$ as:

$\card{B_1 \cap B_2} = \max \set{\card{C_1 \cap C_2} : C_1, C_2 \in \mathscr B : \card{C_1} \ne \card{C_2}}$

It follows that:

$\forall B_1, B_2 \in \mathscr B : \card{B_1} = \card{B_2}$

$\blacksquare$