Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Base Axiom/Definition 1 Iff Definition 4

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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 definitions of the concept of Matroid Base Axiom are equivalent:

Definition 1

$\mathscr B$ is said to satisfy the base axiom if and only if:

\((\text B 1)\)   $:$     \(\displaystyle \forall B_1, B_2 \in \mathscr B:\) \(\displaystyle x \in B_1 \setminus B_2 \implies \exists y \in B_2 \setminus B_1 : \paren {B_1 \setminus \set x} \cup \set y \in \mathscr B \)             


Definition 4

$\mathscr B$ is said to satisfy the base axiom if and only if:

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


Proof

Necessary Condition

Let $\mathscr B$ satisfy the base axiom:

\((\text B 1)\)   $:$     \(\displaystyle \forall B_1, B_2 \in \mathscr B:\) \(\displaystyle x \in B_1 \setminus B_2 \implies \exists y \in B_2 \setminus B_1 : \paren {B_1 \setminus \set x} \cup \set y \in \mathscr B \)             


From Leigh.Samphier/Sandbox/Matroid Satisfies 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 a 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:

\(\displaystyle y\) \(\in\) \(\displaystyle \map C {x, B_2} \setminus B_1\)
\(\displaystyle \) \(\subseteq\) \(\displaystyle \map C {x, B_2} \setminus \set x\) Set Difference with Subset is Superset of Set Difference
\(\displaystyle \) \(\subseteq\) \(\displaystyle \paren{B_2 \cup \set x} \setminus \set x\) Set Difference over Subset
\(\displaystyle \) \(\subseteq\) \(\displaystyle B_2 \setminus \set x\) Set Difference with Union is Set Difference
\(\displaystyle \) \(\subseteq\) \(\displaystyle B_2\) Set Difference is Subset

From Leigh.Samphier/Sandbox/Matroid Base Substitution From Fundamental Circuit:

$\paren{B_2 \setminus \set y} \cup \set x \in \mathscr B$


It follows that $\mathscr B$ satisfies the base axiom:

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

$\Box$


Sufficient Condition

Follows immediately from Definition 4 and Definition 1.

$\blacksquare$