Leigh.Samphier/Sandbox/Matroid Satisfies Base Axiom

Theorem

Let $S$ be a finite set.

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

Then $\mathscr B$ is the set of bases of a matroid on $S$ if and only if $\mathscr B$ 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

Necessary Condition

Let $\mathscr B$ be the set of bases of the matroid on $M = \struct{S, \mathscr I}$

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

Let $x \in B_1 \setminus B_2$.

We have:

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

We have:

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

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

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

$\Box$

Sufficient Condition

Let $\mathscr B$ satisfies the base axiom:

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

Let $\mathscr I = \set{X \subseteq S : \exists B \in \mathscr B : X \subseteq B}$

It is to be shown that:

and

• $\quad \mathscr B$ is the set of bases of the matroid $M = \struct{S, \mathscr I}$

Matroid Axioms

Matroid Axiom $(\text I 1)$

We have $\mathscr B$ is non-empty.

Let $B \in \mathscr B$.

$\O \subseteq B$

By definition of $\mathscr I$:

$\O \in \mathscr I$

It follows that $\mathscr I$ satisfies the matroid axiom $(\text I 1)$ by definition.

$\Box$

Matroid Axiom $(\text I 2)$

Let $X \in \mathscr I$.

Let $Y \subseteq X$.

By definition of $\mathscr I$:

$\exists B \in \mathscr B : X \subseteq B$
$Y \subseteq B$

By definition of $\mathscr I$:

$Y \in \mathscr I$

It follows that $\mathscr I$ satisfies the matroid axiom $(\text I 2)$ by definition.

$\Box$

Matroid Axiom $(\text I 3)$

Let $U, V \in \mathscr I$ such that:

$\card V < \card U$

By definition of $\mathscr I$:

$\exists B_1, B_2 \in \mathscr B : U \subseteq B_1, V \subseteq B_2$
$\exists B_1, B_2 \in \mathscr B : U \subseteq B_1, V \subseteq B_2 : \card{B_1 \cap B_2} = \max \set{\card{C_1 \cap C_2} : U \subseteq C_1, V \subseteq C_2 \text{ and } C_1, C_2 \in \mathscr B}$

$B_2 \cap \paren{U \setminus V} = \O$
Lemma
$\exists B_3 \in \mathscr B$:
$V \subseteq B_3$
$\card{B_1 \cap B_3} > \card{B_1 \cap B_2}$

$\Box$

This contradicts the choice of $B_1, B_2$ such that:

$\card{B_1 \cap B_2} = \max \set{\card{C_1 \cap C_2} : U \subseteq C_1, V \subseteq C_2 \text{ and } C_1, C_2 \in \mathscr B}$

Hence:

$B_2 \cap \paren{U \setminus V} \ne \O$

Let $x \in B_2 \cap \paren{U \setminus V}$.

$V \cup \set x \subseteq B_2$

By definition of $\mathscr I$:

$V \cup \set x \in \mathscr I$

It follows that $\mathscr I$ satisfies the matroid axiom $(\text I 3)$ by definition.

$\Box$

This completes the proof that $M = \struct{S, I}$ forms a matroid.

$\Box$

$\mathscr B$ is Set of Bases

Let $B \in \mathscr B$.

$B \in \mathscr I$

Let $U \in \mathscr I$ such that:

$B \subseteq U$

By definition of $\mathscr I$:

$\exists B' \in \mathscr B : I \subseteq B'$
$B \subseteq B'$
$\card B = \card {B'}$
$B = B'$

By definition of set equality:

$U = B$

It has been shown that $B$ is a maximal subset of the ordered set $\struct{\mathscr I, \subseteq}$.

It follows that $\mathscr B$ is the set of bases of the matroid $M = \struct{S, I}$ by definition.

$\blacksquare$