User:Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Base Axiom/Definition 3 Iff Definition 7

Theorem
Let $S$ be a finite set.

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

Definition 3
$\mathscr B$ is said to satisfy the base axiom :

Definition 7
$\mathscr B$ is said to satisfy the base axiom :

Necessary Condition
Let $\mathscr B$ satisfy the base axiom:

Let $B_1, B_2 \in \mathscr B$.

From $(\text B 3)$:
 * $\exists \text{ a bijection } \pi : B_2 \setminus B_1 \to B_1 \setminus B_2 : \forall y \in B_2 \setminus B_1 : \paren {B_2 \setminus \set y } \cup \set {\map \pi y} \in \mathscr B$

Let $\pi^{-1} : B_1 \setminus B_2 \to B_2 \setminus B_1$ denote the inverse mapping of $\pi$.

From Inverse of Bijection is Bijection, $\pi^{-1}$ is a bijection.

From Inverse Element of Bijection:
 * $\forall x \in B_1 \setminus B_2 : \map {\pi^{-1}} x = y \iff \map \pi y = x$

Hence:
 * $\forall x \in B_1 \setminus B_2 : \paren {B_1 \setminus \set{\map {\pi^{-1}} x} } \cup \set x = \paren {B_1 \setminus \set y} \cup \set {\map \pi y} \in \mathscr B$

It follows that $\mathscr B$ satisfies the base axiom:

Sufficient Condition
Let $\mathscr B$ satisfy the base axiom:

Let $B_1, B_2 \in \mathscr B$.

From $(\text B 7)$:
 * $\exists \text{ a bijection } \pi : B_2 \setminus B_1 \to B_1 \setminus B_2 : \forall y \in B_2 \setminus B_1 : \paren {B_1 \setminus \set {\map \pi y} } \cup \set y \in \mathscr B$

Let $\pi^{-1} : B_1 \setminus B_2 \to B_2 \setminus B_1$ denote the inverse mapping of $\pi$.

From Inverse of Bijection is Bijection, $\pi^{-1}$ is a bijection.

From Inverse Element of Bijection:
 * $\forall x \in B_1 \setminus B_2 : \map {\pi^{-1}} x = y \iff \map \pi y = x$

Hence:
 * $\forall x \in B_1 \setminus B_2 : \paren {B_1 \setminus \set x } \cup \set{\map {\pi^{-1}} x} = \paren {B_1 \setminus \set {\map \pi y} } \cup \set y \in \mathscr B$

It follows that $\mathscr B$ satisfies the base axiom: