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:

\((\text B 1)\)

$:$

\(\displaystyle \forall B_1, B_2 \in \mathscr B:\)

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

From Leigh.Samphier/Sandbox/Matroid Satisfies Base Axiom:

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 $x \in B_1 \setminus B_2$.

From Matroid Base Union External Element has Fundamental Circuit:

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:

$x \in B_1 \cap \map C {x, B_2}$

From Element of Matroid Base and Circuit has a Substitute:

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

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

From Leigh.Samphier/Sandbox/Matroid Base Substitution From Fundamental Circuit:

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

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.

$\blacksquare$