User:Leigh.Samphier/Sandbox/Matroid Satisfies Rank Axioms

Theorem
Let $S$ be a finite set.

Let $\rho : \powerset S \to \Z$ be the rank function of a matroid on $S$.

Then $\rho$ satisfies the rank axioms:

Proof
Let $\rho$ be the rank function of a matroid $M = \struct{S, \mathscr I}$ on $S$

$\rho$ satisfies $(\text R 1)$
$\rho$ satisfies $(\text R 1)$ follows immediately from Rank of Empty Set is Zero.

$\rho$ satisfies $(\text R 2)$
Let:
 * $X \subseteq S$
 * $y \in S$

From Rank Function is Increasing:
 * $\map \rho X \le \map \rho {X \cup \set y}$

Let $Y$ be a maximal independent subset of $X \cup \set y$.

From User:Leigh.Samphier/Sandbox/Cardinality of Maximal Independent Subset Equals Rank of Set
 * $\card Y = \map \rho {X \cup \set y}$

By matroid axiom $(\text I 2)$:
 * $Y \setminus \set y \in \mathscr I$

We now consider two cases.

Case 1: $y \in Y$
We have:

Case 2: $y \notin Y$
We have:

In either case:
 * $\map \rho X + 1 \ge \card Y = \map \rho {X \cup \set y}$

$\rho$ satisfies $(\text R 3)$
Let:
 * $X \subseteq S$
 * $y, z \in S$

Case 1: $y, z \notin X$
Let $A = X \cup \set y \cup \set z$.

From Rank Function is Increasing:
 * $\map \rho A \ge \map \rho X$

We now prove the contrapositive statement:
 * $\map \rho A > \map \rho X \implies$ either $\map \rho {X \cup \set y} > \map \rho X$ or $\map \rho {X \cup \set z} > \map \rho X$

Let $\map \rho A > \map \rho X$.

Let $Y$ be a maximal independent subset of $X$.

Let $Z$ be a maximal independent subset of $A$.

From User:Leigh.Samphier/Sandbox/Cardinality of Maximal Independent Subset Equals Rank of Set:
 * $\card Y = \map \rho X < \map \rho A = \card Z$

From Independent Set can be Augmented by Larger Independent Set there exists non-empty set $Z' \subseteq Z \setminus Y$:
 * $Y \cup Z' \in \mathscr I$
 * $\card{Y \cup Z'} = \card Z$


 * $y, z \notin Z'$
 * $y, z \notin Z'$

We have:

Hence:

This is a contradiction.

Hence:
 * either $y \in Z'$ or $z \in Z'$

, assume $y \in Z'$.

From matroid axiom $(\text I 2)$:
 * $Y \cup \set y \in \mathscr I$

We have:

This proves the contrapositive statement.

Case 2: Either $y \in X$ or $z \in X$
, assume $y \in X$.

Let:
 * $\map \rho {X \cup \set y} = \map \rho {X \cup \set z} = \map \rho X$.

We have:
 * $X \cup \set y \cup \set z = X \cup \set z$

Hence:
 * $\map \rho {X \cup \set y \cup \set z} = \map \rho {X \cup \set z} = \map \rho X$

In either case:
 * $\map \rho {X \cup \set y} = \map \rho {X \cup \set z} = \map \rho X \implies \map \rho {X \cup \set y \cup \set z} = \map \rho X$