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:

\((\text B 4)\)   $:$     \(\ds \forall B_1, B_2 \in \mathscr B:\) \(\ds x \in B_1 \setminus B_2 \implies \exists y \in B_2 \setminus B_1 : \paren {B_1 \setminus \set x} \cup \set y, \paren {B_2 \setminus \set y} \cup \set x \in \mathscr B \)      


$\blacksquare$