# Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)

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## Definition

Let $S$ be a finite set.

Let $\mathscr B$ be a non-empty set of subsets of $S$.

### Definition 1

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

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

### Definition 2

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

 $(\text B 2)$ $:$ $\ds \forall B_1, B_2 \in \mathscr B:$ $\ds x \in B_1 \setminus B_2 \implies \exists y \in B_2 \setminus B_1 : \paren {B_1 \cup \set y} \setminus \set x \in \mathscr B$

### Definition 3

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

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

### Definition 4

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

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

### Definition 5

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

 $(\text B 5)$ $:$ $\ds \forall B_1, B_2 \in \mathscr B:$ $\ds x \in B_1 \setminus B_2 \implies \exists y \in B_2 \setminus B_1 : \paren {B_2 \setminus \set y} \cup \set x \in \mathscr B$

### Definition 6

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

 $(\text B 6)$ $:$ $\ds \forall B_1, B_2 \in \mathscr B:$ $\ds x \in B_1 \setminus B_2 \implies \exists y \in B_2 \setminus B_1 : \paren {B_2 \cup \set x} \setminus \set y \in \mathscr B$

### Definition 7

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

 $(\text B 7)$ $:$ $\ds \forall B_1, B_2 \in \mathscr B:$ $\ds \exists \text{ a bijection } \pi : B_1 \setminus B_2 \to B_2 \setminus B_1 : \forall x \in B_1 \setminus B_2 : \paren {B_2 \setminus \set {\map \pi x} } \cup \set x \in \mathscr B$