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

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:

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

Let $x \in B_1 \setminus B_2$.

We have:

Let $X = B_1 \setminus \set x$.

From Independent Set can be Augmented by Larger Independent Set:
 * $\exists Y \subseteq B_2 \setminus X : X \cup Y \in \mathscr I : \size{X \cup Y} = \size {B_2}$

From Subset of Set Difference iff Disjoint Set:
 * $X \cap Y = \O$

We have:

From Cardinality of Singleton:
 * $\exists y : Y = \set y$

By definition of a subset:
 * $y \in B_2 \setminus X$

By assummption $x \notin B_2$, so:
 * $y \ne x$

Then:
 * $y \in B_2 \setminus B_1$

From choice of $Y$:
 * $X \cup Y \in \mathscr I$

That is:
 * $\paren{ B_1 \setminus \set x} \cup \set y \in \mathscr I$

We have:

Then:
 * $\paren { B_1 \cup \set y} \setminus \set x \in \mathscr I$

We have:

From Independent Subset is Base if Cardinality Equals Rank of Matroid:
 * $\paren { B_1 \cup \set y} \setminus \set x \in \mathscr B$

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