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

\((\text B 1)\)

$:$

\(\ds \forall B_1, B_2 \in \mathscr B:\)

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

if and only if

$\mathscr B$ 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 $1$.

$\blacksquare$