User:Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Rank Axioms/Condition 1 Implies Condition 3

Theorem
Let $S$ be a finite set.

Let $\rho : \powerset S \to \Z$ be a mapping from the power set of $S$ to the integers.

Let $\rho$ satisfy definition 1 of the rank axioms:

Then $\rho$ is the rank function of a matroid on $S$.

Lemma 3
Let:
 * $\mathscr I = \set{X \subseteq S : \map \rho X = \card X}$

It is to be shown that:
 * $\quad \mathscr I$ satisfies the matroid axioms

and
 * $\quad \rho$ is the rank function of the matroid $M = \struct{S, \mathscr I}$

Matroid Axiom $(\text I 1)$
We have:

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

Hence:
 * $M$ satisfies matroid axiom $(\text I 1)$.

Matroid Axiom $(\text I 2)$
Let
 * $X \in \mathscr I$


 * $\exists Y \subseteq X : Y \notin \mathscr I$

Let:
 * $Y_0 \subseteq X : \card {Y_0} = \max \set{\card Z : Z \subseteq X \land Z \notin \mathscr I}$

By definition of $\mathscr I$:
 * $Y_0 \notin \mathscr I \leadsto \map \rho {Y_0} \ne \card {Y_0}$

From Lemma 2:
 * $\map \rho {Y_0} < \card {Y_0}$

As $X \in \mathscr I$ then:
 * $Y_0 \ne X$

From Set Difference with Proper Subset:
 * $X \setminus Y_0 \ne \O$

Let $y \in X \setminus Y_0$.

We have:

This is a contradiction.

So:
 * $\forall Y \subseteq X : Y \in \mathscr I$

Hence:
 * $M$ satisfies matroid axiom $(\text I 2)$.

Matroid Axiom $(\text I 3)$
Let
 * $U \in \mathscr I$
 * $V \subseteq S$
 * $\card U < \card V$

We prove the contrapositive statement:
 * $\paren{\forall x \in V \setminus U : U \cup \set x \notin \mathscr I} \implies V \notin \mathscr I$

Let for all $x \in V \setminus U$: $U \cup \set x \notin \mathscr I$

That is:
 * $\forall x \in V \setminus U : \map \rho {U \cup \set x} \ne \card {U \cup \set x}$

We have:

Hence:
 * $V \notin \mathscr I$

$\rho$ is Rank Function
Let $\rho_M$ be the rank function of the matroid $M = \struct{S, \mathscr I}$.

Let $X \subseteq S$.

By definition of the rank function:
 * $\map {\rho_M} X = \max \set{\card Y : Y \subseteq X, Y \in \mathscr I}$

Let $Y_0 \subseteq X$:
 * $\card {Y_0} = \max \set{\card Y : Y \subseteq X, Y \in \mathscr I}$

We have:

So it remains to show:
 * $\map \rho {Y_0} = \map \rho X$

Case 1 : $Y_0 = X$
Let $Y_0 = X$.

Then:
 * $\map \rho {Y_0} = \map \rho X$

Case 2 : $Y_0 \ne X$
Let $Y_0 \ne X$.

Then:
 * $Y_0 \subsetneq X$

From Set Difference with Proper Subset:
 * $X \setminus Y_0 \ne \O$

By choice of $Y_0$:
 * $\forall y \in X \setminus Y_0 : Y_0 \cup \set y \notin \mathscr I$

That is:
 * $\forall y \in X \setminus Y_0 : \map \rho {Y_0 \cup \set y} \ne \card {Y_0 \cup \set y}$

From Lemma 3:
 * $\map \rho {Y_0} = \map \rho {Y_0 \cup X} = \map \rho X$

In either case:
 * $\map \rho {Y_0} = \map \rho X$

It follows that:
 * $\forall X \subseteq S : \map {\rho_M} X = \map \rho X$

Hence $\rho$ is the rank function of the matroid $M = \struct{S, \mathscr I}$.