User:Leigh.Samphier/Matroids/Equivalence of Definitions of Matroid Base Axioms

Theorem
Let $S$ be a finite set.

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

Definition 1 iff Definition 2
Definition 1 holds Definition 2 holds follows immediately from the lemma.

Definition 1 implies Definition 3
Let $\mathscr B$ satisfy the base axiom:

Definition 3 implies Definition 1
By choosing $y = \map \pi x$ in Definition 3, Definition 1 follows immediately.

Definition 1 implies Definition 4
Let $\mathscr B$ satisfy the base axiom:

Definition 4 implies Definition 1
Follows immediately from Definition 4 and Definition 1.

Definition 4 implies Definition 5
Follows immediately from Definition 4 and Definition 5.

Definition 5 iff Definition 6
Definition 5 holds Definition 6 holds follows immediately from the lemma.

Definition 5 implies Definition 7
Let $\mathscr B$ satisfy the base axiom:

Definition 7 implies Definition 5
By choosing $y = \map \pi x$ in Definition 7, Definition 5 follows immediately.

Definition 7 implies Definition 3
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:

Definition 7 implies Definition 1
Let $\mathscr B$ satisfy the base axiom: