User:Leigh.Samphier/Matroids/Equivalence of Definitions of Matroid Base Axioms
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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 statements are equivalent:
Formulation 1
$\mathscr B$ is said to satisfy the base axiom if and only if:
| \((\text B 1)\) | $:$ | \(\ds \forall B_1, B_2 \in \mathscr B:\) | \(\ds x \in B_1 \setminus B_2 \implies \exists y \in B_2 \setminus B_1 : \paren {B_1 \setminus \set x} \cup \set y \in \mathscr B \) |
Formulation 2
$\mathscr B$ is said to satisfy the base axiom if and only if:
| \((\text B 2)\) | $:$ | \(\ds \forall B_1, B_2 \in \mathscr B:\) | \(\ds x \in B_1 \setminus B_2 \implies \exists y \in B_2 \setminus B_1 : \paren {B_1 \cup \set y} \setminus \set x \in \mathscr B \) |
Formulation 3
$\mathscr B$ is said to satisfy the base axiom if and only if:
| \((\text B 3)\) | $:$ | \(\ds \forall B_1, B_2 \in \mathscr B:\) | \(\ds \exists \text{ a bijection } \pi : B_1 \setminus B_2 \to B_2 \setminus B_1 : \forall x \in B_1 \setminus B_2 : \paren {B_1 \setminus \set x } \cup \set {\map \pi x} \in \mathscr B \) |
Formulation 4
$\mathscr B$ is said to satisfy the base axiom if and only if:
| \((\text B 4)\) | $:$ | \(\ds \forall B_1, B_2 \in \mathscr B:\) | \(\ds x \in B_1 \setminus B_2 \implies \exists y \in B_2 \setminus B_1 : \paren {B_1 \setminus \set x} \cup \set y, \paren {B_2 \setminus \set y} \cup \set x \in \mathscr B \) |
Formulation 5
$\mathscr B$ is said to satisfy the base axiom if and only if:
| \((\text B 5)\) | $:$ | \(\ds \forall B_1, B_2 \in \mathscr B:\) | \(\ds x \in B_1 \setminus B_2 \implies \exists y \in B_2 \setminus B_1 : \paren {B_2 \setminus \set y} \cup \set x \in \mathscr B \) |
Formulation 6
$\mathscr B$ is said to satisfy the base axiom if and only if:
| \((\text B 6)\) | $:$ | \(\ds \forall B_1, B_2 \in \mathscr B:\) | \(\ds x \in B_1 \setminus B_2 \implies \exists y \in B_2 \setminus B_1 : \paren {B_2 \cup \set x} \setminus \set y \in \mathscr B \) |
Formulation 7
$\mathscr B$ is said to satisfy the base axiom if and only if:
| \((\text B 7)\) | $:$ | \(\ds \forall B_1, B_2 \in \mathscr B:\) | \(\ds \exists \text{ a bijection } \pi : B_1 \setminus B_2 \to B_2 \setminus B_1 : \forall x \in B_1 \setminus B_2 : \paren {B_2 \setminus \set {\map \pi x} } \cup \set x \in \mathscr B \) |
Proof
Formulation $1$ iff Formulation $2$
Formulation $1$ holds if and only if formulation $2$ holds follows immediately from the User:Leigh.Samphier/Matroids/Corollary of Set Difference Then Union Equals Union Then Set Difference.
$\Box$
Formulation $1$ iff Formulation $3$
Necessary Condition
Follows immediately from:
Sufficient Condition
By choosing $y = \map \pi x$ in formulation $3$, formulation $1$ follows immediately.
$\Box$
Formulation $1$ iff Formulation $4$
Necessary Condition
Let $\mathscr B$ satisfy formulation $1$ of base axiom:
| \((\text B 1)\) | $:$ | \(\ds \forall B_1, B_2 \in \mathscr B:\) | \(\ds x \in B_1 \setminus B_2 \implies \exists y \in B_2 \setminus B_1 : \paren {B_1 \setminus \set x} \cup \set y \in \mathscr B \) |
From User:Leigh.Samphier/Matroids/Matroid Bases Iff Satisfies Formulation 1 of Matroid 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 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 Matroid Base Substitution From Fundamental Circuit:
- $\paren{B_2 \setminus \set y} \cup \set x \in \mathscr B$
It follows that $\mathscr B$ satisfies formulation $4$ of base axiom:
| \((\text B 4)\) | $:$ | \(\ds \forall B_1, B_2 \in \mathscr B:\) | \(\ds x \in B_1 \setminus B_2 \implies \exists y \in B_2 \setminus B_1 : \paren {B_1 \setminus \set x} \cup \set y, \paren {B_2 \setminus \set y} \cup \set x \in \mathscr B \) |
Sufficient Condition
Follows immediately from formulation $4$ and formulation $1$.
$\Box$
Formulation $1$ iff Formulation $5$
Follows immediately from:
$\Box$
Formulation $3$ iff Formulation $7$
Necessary Condition
Let $\mathscr B$ satisfy formulation $3$ of 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:
- $\forall x \in B_1 \setminus B_2 : \paren {B_1 \setminus \set{\map {\pi^{-1}} x} } \cup \set x = \paren {B_1 \setminus \set y} \cup \set {\map \pi y} \in \mathscr B$
Let $\mathscr B$ satisfy formulation $7$ of 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 formulation $7$ of 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:
- $\forall x \in B_1 \setminus B_2 : \paren {B_1 \setminus \set x } \cup \set{\map {\pi^{-1}} x} = \paren {B_1 \setminus \set {\map \pi y} } \cup \set y \in \mathscr B$
Let $\mathscr B$ satisfy formulation $3$ of base axiom:
| \((\text B 3)\) | $:$ | \(\ds \forall B_1, B_2 \in \mathscr B:\) | \(\ds \exists \text{ a bijection } \pi : B_1 \setminus B_2 \to B_2 \setminus B_1 : \forall x \in B_1 \setminus B_2 : \paren {B_1 \setminus \set x } \cup \set {\map \pi x} \in \mathscr B \) |
$\Box$
Formulation $5$ iff Formulation $6$
Formulation $5$ holds if and only if formulation $6$ holds follows immediately from the User:Leigh.Samphier/Matroids/Corollary of Set Difference Then Union Equals Union Then Set Difference.
$\blacksquare$