User:Leigh.Samphier/Matroids/Equivalence of Definitions of Matroid Base Axioms/Formulation 1 Iff Formulation 4
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Theorem
Let $S$ be a finite set.
Let $\mathscr B$ be a non-empty set of subsets of $S$.
Then:
- $\mathscr B$ satisfies formulation $1$ of base axiom
\((\text B 1)\) | $:$ | \(\ds \forall B_1, B_2 \in \mathscr B:\) | \(\ds x \in B_1 \setminus B_2 \implies \exists y \in B_2 \setminus B_1 : \paren {B_1 \setminus \set x} \cup \set y \in \mathscr B \) |
- $\mathscr B$ satisfies formulation $4$ of base axiom
\((\text B 4)\) | $:$ | \(\ds \forall B_1, B_2 \in \mathscr B:\) | \(\ds x \in B_1 \setminus B_2 \implies \exists y \in B_2 \setminus B_1 : \paren {B_1 \setminus \set x} \cup \set y, \paren {B_2 \setminus \set y} \cup \set x \in \mathscr B \) |
Proof
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 $1$.
$\blacksquare$