# 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:

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

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.

$\Box$

### 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.

$\forall x \in B_1 \setminus B_2 : \map {\pi^{-1}} x = y \iff \map \pi y = x$

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.

$\forall x \in B_1 \setminus B_2 : \map {\pi^{-1}} x = y \iff \map \pi y = x$

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:

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

$\blacksquare$