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

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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:

\((\text B 1)\)   $:$     \(\ds \forall B_1, B_2 \in \mathscr B:\) \(\ds x \in B_1 \setminus B_2 \implies \exists y \in B_2 \setminus B_1 : \paren {B_1 \setminus \set x} \cup \set y \in \mathscr B \)             


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

Let $x \in B_1 \setminus B_2$.

We have:

\(\ds \size {B_1 \setminus \set x}\) \(=\) \(\ds \size {B_1} - \size {\set x}\) Cardinality of Set Difference with Subset
\(\ds \) \(=\) \(\ds \size {B_2} - \size {\set x}\) All Bases of Matroid have same Cardinality
\(\ds \) \(=\) \(\ds \size {B_2} - 1\) Cardinality of Singleton
\(\ds \) \(<\) \(\ds \size {B_2}\)

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:

\(\ds B_2 \setminus \paren{B_1 \setminus \set x}\) \(=\) \(\ds \paren{B_2 \setminus B_1} \cup \paren{B_2 \cap \set x}\) Set Difference with Set Difference is Union of Set Difference with Intersection
\(\ds \) \(=\) \(\ds \paren{B_2 \setminus B_1} \cup \O\) Intersection With Singleton is Disjoint if Not Element
\(\ds \) \(=\) \(\ds B_2 \setminus B_1\) Union with Empty Set


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

We have:

\(\ds \size{\paren { B_1 \setminus \set x} \cup \set y}\) \(=\) \(\ds \size{B_1 \setminus \set x} + \size{\set y}\) Corollary to Cardinality of Set Union
\(\ds \) \(=\) \(\ds \size{B_1} - \size{\set x} + \size{\set y}\) Cardinality of Set Difference with Subset
\(\ds \) \(=\) \(\ds \size{B_1} - 1 + 1\) Cardinality of Singleton
\(\ds \) \(=\) \(\ds \size{B_1}\)

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.