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

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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 implies Definition 3

Let $\mathscr B$ satisfy the base axiom:

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


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

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

$\Box$


Definition 3 implies Definition 1

Follows immediately from Definition 3 and Definition 1.

$\Box$


Definition 1 implies Definition 4

Let $\mathscr B$ satisfy the base axiom:

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


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

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

$\Box$


Definition 4 implies Definition 5

Follows immediately from Definition 4 and Definition 5.

$\Box$


Definition 5 iff Definition 6

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

$\Box$


Definition 5 implies Definition 7

Let $\mathscr B$ satisfy the base axiom:

\((\text B 5)\)   $:$     \(\displaystyle \forall B_1, B_2 \in \mathscr B:\) \(\displaystyle x \in B_1 \setminus B_2 \implies \exists y \in B_2 \setminus B_1 : \paren {B_2 \setminus \set y} \cup \set x \in \mathscr B \)             

Lemma

$\forall B_1, B_2 \in \mathscr B : \card{B_1} = \card{B_2}$

$\Box$


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

\((\text B 7)\)   $:$     \(\displaystyle \forall B_1, B_2 \in \mathscr B:\) \(\displaystyle \exists \text{ a bijection } \pi : B_1 \setminus B_2 \to B_2 \setminus B_2 : \forall x \in B_1 \setminus B_2 : \paren {B_2 \setminus \set {\map \pi x} } \cup \set x \in \mathscr B \)             

$\Box$


Definition 7 implies Definition 1

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 $x \in B_1 \setminus B_2$.

Let $y = \map{\pi^{-1}} x$.

Then:

$x = \map \pi y$

and:

$\paren {B_1 \setminus \set x } \cup \set y \in \mathscr B$


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

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

$\blacksquare$