# Leigh.Samphier/Sandbox/Matroid Satisfies Base Axiom/Necessary Condition

## Theorem

Let $M = \struct {S, \mathscr I}$ be a matroid.

Let $\mathscr B$ be the set of bases of the matroid on $M$.

Then $\mathscr B$ satisfies the base axiom:

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

## Proof

Let $B_1, B_2 \in \mathscr B$.

Let $x \in B_1 \setminus B_2$.

We have:

 $\displaystyle \size {B_1 \setminus \set x}$ $=$ $\displaystyle \size {B_1} - \size {\set x}$ Cardinality of Set Difference with Subset $\displaystyle$ $=$ $\displaystyle \size {B_2} - \size {\set x}$ All Bases of Matroid have same Cardinality $\displaystyle$ $=$ $\displaystyle \size {B_2} - 1$ Cardinality of Singleton $\displaystyle$ $<$ $\displaystyle \size {B_2}$
$\exists y \in B_2 \setminus \paren{B_1 \setminus \set x} : \paren{ B_1 \setminus \set x} \cup \set y \in \mathscr I$

We have:

 $\displaystyle B_2 \setminus \paren{B_1 \setminus \set x}$ $=$ $\displaystyle \paren{B_2 \setminus B_1} \cup \paren{B_2 \cap \set x}$ Set Difference with Set Difference is Union of Set Difference with Intersection $\displaystyle$ $=$ $\displaystyle \paren{B_2 \setminus B_1} \cup \O$ Intersection With Singleton is Disjoint if Not Element $\displaystyle$ $=$ $\displaystyle B_2 \setminus B_1$ Union with Empty Set

Then:

$\exists y \in B_2 \setminus B_1 : \paren{ B_1 \setminus \set x} \cup \set y \in \mathscr I$

We have:

 $\displaystyle \size{\paren { B_1 \setminus \set x} \cup \set y}$ $=$ $\displaystyle \size{B_1 \setminus \set x} + \size{\set y}$ Corollary to Cardinality of Set Union $\displaystyle$ $=$ $\displaystyle \size{B_1} - \size{\set x} + \size{\set y}$ Cardinality of Set Difference with Subset $\displaystyle$ $=$ $\displaystyle \size{B_1} - 1 + 1$ Cardinality of Singleton $\displaystyle$ $=$ $\displaystyle \size{B_1}$
$\paren { B_1 \setminus \set x} \cup \set y \in \mathscr B$

Since $x$, $B_1$ and $B_2$ were arbitrary then the result follows.

$\blacksquare$