User:Leigh.Samphier/Matroids/Matroid Bases Satisfy Formulation 4 Base Axiom
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Theorem
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $\mathscr B$ be the set of bases of the matroid $M$.
Then $\mathscr B$ 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
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:
\((\text B 4)\) | $:$ | \(\ds \forall B_1, B_2 \in \mathscr B:\) | \(\ds x \in B_1 \setminus B_2 \implies \exists y \in B_2 \setminus B_1 : \paren {B_1 \setminus \set x} \cup \set y, \paren {B_2 \setminus \set y} \cup \set x \in \mathscr B \) |
$\blacksquare$