User:Leigh.Samphier/Matroids/Equivalence of Definitions of Matroid Base Axioms/Axiom B3 Iff Axiom B7
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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 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:
\((\text B 3)\) | $:$ | \(\ds \forall B_1, B_2 \in \mathscr B:\) | \(\ds \exists \text{ a bijection } \pi : B_1 \setminus B_2 \to B_2 \setminus B_1 : \forall x \in B_1 \setminus B_2 : \paren {B_1 \setminus \set x } \cup \set {\map \pi x} \in \mathscr B \) |
$\blacksquare$