Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Base Axiom/Definition 1 Iff Definition 3
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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 3
$\mathscr B$ is said to satisfy the base axiom if and only if:
\((\text B 3)\) | $:$ | \(\displaystyle \forall B_1, B_2 \in \mathscr B:\) | \(\displaystyle \exists \text{ a bijection } \pi : B_1 \setminus B_2 \to B_2 \setminus B_1 : \forall x \in B_1 \setminus B_2 : \paren {B_1 \setminus \set x } \cup \set {\map \pi x} \in \mathscr B \) |
Proof
Necessary Condition
Let $\mathscr B$ satisfy the base axiom:
\((\text B 1)\) | $:$ | \(\displaystyle \forall B_1, B_2 \in \mathscr B:\) | \(\displaystyle x \in B_1 \setminus B_2 \implies \exists y \in B_2 \setminus B_1 : \paren {B_1 \setminus \set x} \cup \set y \in \mathscr B \) |
$\dots$
It follows that $\mathscr B$ satisfies the base axiom:
\((\text B 3)\) | $:$ | \(\displaystyle \forall B_1, B_2 \in \mathscr B:\) | \(\displaystyle \exists \text{ a bijection } \pi : B_1 \setminus B_2 \to B_2 \setminus B_1 : \forall x \in B_1 \setminus B_2 : \paren {B_1 \setminus \set x } \cup \set {\map \pi x} \in \mathscr B \) |
$\Box$
Sufficient Condition
By choosing $y = \map \pi x$ in Definition 3, Definition 1 follows immediately.
$\blacksquare$