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

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:

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

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

and from Max Operation Equals an Operand let:
 * $\card{B_1 \cap B_2} = \max \set{\card{C_1 \cap C_2} : C_1, C_2 \in \mathscr B : \card{C_1} \ne \card{C_2}}$

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:

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