Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Base Axiom/Definition 1 Iff Definition 3

From ProofWiki
Jump to navigation Jump to search

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.

$\blacksquare$