User:Leigh.Samphier/Matroids/Equivalence of Definitions of Matroid Base Axioms/Axiom B3 Iff Axiom B7

This page needs proofreading. Please check it for mathematical errors. If you believe there are none, please remove {{Proofread}} from the code. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Proofread}} from the code.

Theorem

Let $S$ be a finite set.

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

The following definitions for the Matroid Base Axioms are equivalent:

Axiom $(\text B 3)$

\((\text B 3)\)

$:$

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

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

Axiom $(\text B 7)$

\((\text B 7)\)

$:$

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

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

Proof

Necessary Condition

Let $\mathscr B$ satisfy the base axiom:

\((\text B 3)\)

$:$

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

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

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:

\((\text B 7)\)

$:$

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

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

Sufficient Condition

Let $\mathscr B$ satisfy the base axiom:

\((\text B 7)\)

$:$

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

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

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 inyverse 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:

\((\text B 3)\)

$:$

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

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

$\blacksquare$