User:Leigh.Samphier/Matroids/Equivalence of Definitions of Matroid Base Axioms/Axiom B1 Implies Set of Matroid Bases
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Theorem
Let $S$ be a finite set.
Let $\mathscr B$ be a non-empty set of subsets of $S$ satisfying the base axiom: User:Leigh.Samphier/Matroids/Axiom:Base Axioms (Matroid)/Axiom 1
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
Matroid Axioms
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.
$\Box$
Matroid Axiom $(\text I 2)$
Let $X \in \mathscr I$.
Let $Y \subseteq X$.
By definition of $\mathscr I$:
- $\exists B \in \mathscr B : X \subseteq B$
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.
$\Box$
Matroid Axiom $(\text I 3)$
Let $U, V \in \mathscr I$ such that:
- $\card V < \card U$
By definition of $\mathscr I$:
- $\exists B_1, B_2 \in \mathscr B : U \subseteq B_1, V \subseteq B_2$
From Max Operation Equals an Operand:
- $\exists B_1, B_2 \in \mathscr B : U \subseteq B_1, V \subseteq B_2 : \card{B_1 \cap B_2} = \max \set{\card{C_1 \cap C_2} : U \subseteq C_1, V \subseteq C_2 \text{ and } C_1, C_2 \in \mathscr B}$
Aiming for a contradiction, suppose:
- $B_2 \cap \paren{U \setminus V} = \O$
Lemma 1
- $\exists B_3 \in \mathscr B$:
- $V \subseteq B_3$
- $\card{B_1 \cap B_3} > \card{B_1 \cap B_2}$
$\Box$
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 $x \in B_2 \cap \paren{U \setminus V}$.
From Union of Subsets is Subset:
- $V \cup \set x \subseteq B_2$
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 $B \in \mathscr B$.
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:
- $B \subseteq B'$
Lemma 2
- $\forall B_1, B_2 \in \mathscr B : \card{B_1} = \card{B_2}$
where $\card{B_1}$ and $\card{B_2}$ denote the cardinality of the sets $B_1$ and $B_2$ respectively.
$\Box$
From Lemma 2:
- $\card B = \card {B'}$
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.
$\blacksquare$