Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Base Axiom/Definition 1 Iff Definition 4

Theorem

Let $S$ be a finite set.

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

The following definitions of the concept of Matroid Base Axiom are equivalent:

Definition 1

$\mathscr B$ is said to satisfy the base axiom if and only if:

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

Definition 4

$\mathscr B$ is said to satisfy the base axiom if and only if:

 $(\text B 4)$ $:$ $\displaystyle \forall B_1, B_2 \in \mathscr B:$ $\displaystyle x \in B_1 \setminus B_2 \implies \exists y \in B_2 \setminus B_1 : \paren {B_1 \setminus \set x} \cup \set y, \paren {B_2 \setminus \set y} \cup \set x \in \mathscr B$

Proof

Necessary Condition

Let $\mathscr B$ satisfy the base axiom:

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

there exists a matroid $M = \struct{S, \mathscr I}$ such that $\mathscr B$ is the set of bases of $M$.

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

Let $x \in B_1 \setminus B_2$.

there exists a fundamental circuit $\map C {x, B_2}$ of $M$ such that $x \in \map C {x, B_2} \subseteq B_2 \cup \set x$

By definition of set intersection:

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

We have:

 $\ds y$ $\in$ $\ds \map C {x, B_2} \setminus B_1$ $\ds$ $\subseteq$ $\ds \map C {x, B_2} \setminus \set x$ Set Difference with Subset is Superset of Set Difference $\ds$ $\subseteq$ $\ds \paren{B_2 \cup \set x} \setminus \set x$ Set Difference over Subset $\ds$ $\subseteq$ $\ds B_2 \setminus \set x$ Set Difference with Union is Set Difference $\ds$ $\subseteq$ $\ds B_2$ Set Difference is Subset
$\paren{B_2 \setminus \set y} \cup \set x \in \mathscr B$

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

 $(\text B 4)$ $:$ $\displaystyle \forall B_1, B_2 \in \mathscr B:$ $\displaystyle x \in B_1 \setminus B_2 \implies \exists y \in B_2 \setminus B_1 : \paren {B_1 \setminus \set x} \cup \set y, \paren {B_2 \setminus \set y} \cup \set x \in \mathscr B$

$\Box$

Sufficient Condition

Follows immediately from Definition 4 and Definition 1.

$\blacksquare$