User:Leigh.Samphier/Matroids/Equivalence of Definitions of Matroid Base Axioms/Axiom B1 Iff Axiom B3

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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 statements are equivalent:

Axiom $(\text B 1)$

$\mathscr B$ satisfies the base axiom:

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

Axiom $(\text B 3)$

$\mathscr B$ satisfies the base axiom:

\((\text B 3)\)   $:$     \(\ds \forall B_1, B_2 \in \mathscr B:\) \(\ds \exists \text{ a bijection } \pi : B_1 \setminus B_2 \to B_2 \setminus B_1 : \forall x \in B_1 \setminus B_2 : \paren {B_1 \setminus \set x } \cup \set {\map \pi x} \in \mathscr B \)      


Proof

Necessary Condition

Let $\mathscr B$ satisfy the base axiom:

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


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

\((\text B 3)\)   $:$     \(\ds \forall B_1, B_2 \in \mathscr B:\) \(\ds \exists \text{ a bijection } \pi : B_1 \setminus B_2 \to B_2 \setminus B_1 : \forall x \in B_1 \setminus B_2 : \paren {B_1 \setminus \set x } \cup \set {\map \pi x} \in \mathscr B \)      

Sufficient Condition

By choosing $y = \map \pi x$ in Axiom $(\text B 3)$, Axiom $(\text B 1)$ follows immediately.

$\blacksquare$

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