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 that 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 $\mathscr I$:
 * $\O \in \mathscr I$

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 $\mathscr I$:
 * $Y \in \mathscr I$

Matroid Axiom $(\text I 3)$
Let $U, V \in \mathscr I$ such that:
 * $\size U = \size V + 1$

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

Let:

Consider $b_1$.

By base axiom $(\text B 1)$:
 * $\exists z_1 \in B_2 \setminus B_1 : \paren{B_1 \cup \set {z_1}} \setminus \set {b_1} \in \mathscr B$

By definition of $\mathscr I$:
 * $X \cup \set {z_1} \in \mathscr I$

If $z_1 \in Y$ then the matroid axiom $(\text I 3)$ is satisfied.

If $z_1 \notin Y$ then consider $b_2$.

By base axiom $(\text B 1)$:
 * $\exists z_2 \in B_2 \setminus \paren{\paren{B_1 \cup \set {z_1}} \setminus \set {b_1}} : \paren{\paren{B_1 \cup \set {z_1}} \setminus \set {b_1} \cup \set {z_2}} \setminus \set {b_2} \in \mathscr B$

By definition of $\mathscr I$:
 * $X \cup \set {z_2} \in \mathscr I$

If $z_2 \in Y$ then the matroid axiom $(\text I 3)$ is satisfied.

If $z_2 \notin Y$ then replace $b_3$ and so on.

Since $\size{\set{b_1, b_2, \dots, b_q}} > \size{\set{c_1, c_2, \dots , c_{q - 1}}}$, after at most $q$ steps we must replace $b_i$ by a element of $Y$.