Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Base Axiom

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 2

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

\((\text B 2)\)

$:$

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

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

Definition 3

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

\((\text B 3)\)

$:$

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

\(\displaystyle \exists \text{ a bijection } \pi : B_1 \setminus B_2 \to B_2 \setminus B_1 : \forall x \in B_1 \setminus B_2 : \paren {B_1 \setminus \set x } \cup \set {\map \pi x} \in \mathscr B \)

Definition 4

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

\((\text B 4)\)

$:$

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

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

Definition 5

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

\((\text B 5)\)

$:$

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

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

Definition 6

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

\((\text B 6)\)

$:$

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

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

Definition 7

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

\((\text B 7)\)

$:$

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

\(\displaystyle \exists \text{ a bijection } \pi : B_1 \setminus B_2 \to B_2 \setminus B_1 : \forall x \in B_1 \setminus B_2 : \paren {B_2 \setminus \set {\map \pi x} } \cup \set x \in \mathscr B \)

Proof

Lemma

Let $B_1, B_2 \subseteq S$.

Let $x \in B_1 \setminus B_2$.

Let $y \in B_2 \setminus B_1$.

Then:

$\paren{B_1 \setminus \set x} \cup \set y = \paren{B_1 \cup \set y} \setminus \set x$

$\Box$

Definition 1 iff Definition 2

Definition 1 holds if and only if Definition 2 holds follows immediately from the lemma.

$\Box$

Definition 1 iff Definition 3

Necessary Condition

Let $\mathscr B$ satisfy the base axiom:

\((\text B 1)\)

$:$

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

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

$\dots$

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

\((\text B 3)\)

$:$

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

\(\displaystyle \exists \text{ a bijection } \pi : B_1 \setminus B_2 \to B_2 \setminus B_1 : \forall x \in B_1 \setminus B_2 : \paren {B_1 \setminus \set x } \cup \set {\map \pi x} \in \mathscr B \)

$\Box$

Sufficient Condition

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

$\Box$

Definition 1 iff Definition 4

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.

$\Box$

Definition 4 iff Definition 5

Necessary Condition

Follows immediately from Definition 4 and Definition 5.

$\Box$

Sufficient Condition

Let $\mathscr B$ satisfy the base axiom:

\((\text B 5)\)

$:$

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

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

$\Box$

Definition 5 iff Definition 6

Definition 5 holds if and only if Definition 6 holds follows immediately from the lemma.

$\Box$

Definition 3 iff Definition 7

Necessary Condition

Let $\mathscr B$ satisfy the base axiom:

\((\text B 3)\)

$:$

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

\(\displaystyle \exists \text{ a bijection } \pi : B_1 \setminus B_2 \to B_2 \setminus B_1 : \forall x \in B_1 \setminus B_2 : \paren {B_1 \setminus \set x } \cup \set {\map \pi x} \in \mathscr B \)

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:

\((\text B 7)\)

$:$

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

\(\displaystyle \exists \text{ a bijection } \pi : B_1 \setminus B_2 \to B_2 \setminus B_1 : \forall x \in B_1 \setminus B_2 : \paren {B_2 \setminus \set {\map \pi x} } \cup \set x \in \mathscr B \)

$\Box$

Sufficient Condition

Let $\mathscr B$ satisfy the base axiom:

\((\text B 7)\)

$:$

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

\(\displaystyle \exists \text{ a bijection } \pi : B_1 \setminus B_2 \to B_2 \setminus B_1 : \forall x \in B_1 \setminus B_2 : \paren {B_2 \setminus \set {\map \pi x} } \cup \set x \in \mathscr B \)

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:

\((\text B 3)\)

$:$

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

\(\displaystyle \exists \text{ a bijection } \pi : B_1 \setminus B_2 \to B_2 \setminus B_1 : \forall x \in B_1 \setminus B_2 : \paren {B_1 \setminus \set x } \cup \set {\map \pi x} \in \mathscr B \)

$\blacksquare$