User:Leigh.Samphier/Matroids/Equivalence of Definitions of Matroid Base Axioms/Axiom B1 Implies Set of Matroid Bases

Theorem
Let $S$ be a finite set.

Let $\mathscr B$ be a non-empty set of subsets of $S$ satisfying the base axiom:

Then $\mathscr B$ is the set of bases of a matroid on $S$.

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

It is to be shown that:
 * $\quad \mathscr I$ satisfies the matroid axioms

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

Matroid Axiom $(\text I 1)$
We have $\mathscr B$ is non-empty.

Let $B \in \mathscr B$.

From Empty Set is Subset of All Sets:
 * $\O \subseteq B$

By definition of $\mathscr I$:
 * $\O \in \mathscr I$

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

Matroid Axiom $(\text I 2)$
Let $X \in \mathscr I$.

Let $Y \subseteq X$.

By definition of $\mathscr I$:
 * $\exists B \in \mathscr B : X \subseteq B$

From Subset Relation is Transitive:
 * $Y \subseteq B$

By definition of $\mathscr I$:
 * $Y \in \mathscr I$

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

Matroid Axiom $(\text I 3)$
Let $U, V \in \mathscr I$ such that:
 * $\card V < \card U$

By definition of $\mathscr I$:
 * $\exists B_1, B_2 \in \mathscr B : U \subseteq B_1, V \subseteq B_2$

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


 * $B_2 \cap \paren{U \setminus V} = \O$
 * $B_2 \cap \paren{U \setminus V} = \O$

Lemma
From Subset of Set Difference iff Disjoint Set:
 * $U \setminus V \subseteq B_1 \setminus B_2$

We have:

and

We have:

Hence: $\exists y \in \paren{B_2 \setminus B_1} \setminus V$

From $(\text B 1)$:
 * $\exists z \in B_1 \setminus B_2 : \paren{B_2 \setminus \set y} \cup \set z \in \mathscr B$

We have:

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

From Union of Subsets is Subset:
 * $V \cup \set x \subseteq B2$

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.

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

$\mathscr B$ is Set of Bases
Let $B \in \mathscr B$.

From Set is Subset of Itself:
 * $B \in \mathscr I$

Let $U \in \mathscr I$ such that:
 * $B \subseteq U$

By definition of $\mathscr I$:
 * $\exists B' \in \mathscr B : I \subseteq B'$

From Subset Relation is Transitive:
 * $B \subseteq B'$

From Leigh.Samphier/Sandbox/Matroid Base Axiom Implies Sets Have Same Cardinality:
 * $\card B = \card {B'}$

From Cardinality of Proper Subset of Finite Set:
 * $B = B'$

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.