User:Leigh.Samphier/Sandbox/Matroid Satisfies Base Axiom/Necessary Condition

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:

By matroid axiom $(\text I 3)$:
 * $\exists y \in B_2 \setminus \paren{B_1 \setminus \set x} : \paren{ B_1 \setminus \set x} \cup \set y \in \mathscr I$

We have:

Then:
 * $\exists y \in B_2 \setminus B_1 : \paren{ B_1 \setminus \set x} \cup \set y \in \mathscr I$

We have:

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

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