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 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 $(\text I 1)$
From Cardinality of Empty Set:
 * $\card \O = 0$

By rank axiom $(\text R 1)$:
 * $\map \rho \O = 0$

Hence:
 * $\map \rho \O = \card \O$

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 $\O, X \in \mathscr I$ then:
 * $Y_0 \ne \O$
 * $Y_0 \ne X$

From Set Difference with Proper Subset is Proper Subset
 * $X \setminus Y_0 \neq \O$

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:
 * $V \setminus U = \set{x_1, x_2, \dots, x_k}$

Let:
 * $\forall i \in \closedint 1 k : U \cup \set {x_i} \notin \mathscr I$

For each $i \in \closedint 1 k$, let:
 * $V_i = \begin{cases}

U & : i = 1 \\ U \cup \set{x_1, x_2, \dots, x_{i-1}} & : i \in \closedint 2 k \end{cases}$


 * $\mathscr A_i = \set{V_i \cup \set y : y \in V \setminus V_i}$

We note that:

We prove by induction:
 * $\forall i \in \closedint 1 k : \forall Y \in \mathscr A_i : \map \rho Y = \map \rho U$

For all $i \in \closedint 1 k$, let $\map P i$ be the proposition:
 * $\forall Y \in \mathscr A_i : \map \rho Y = \map \rho U$

Basis for the Induction
We have:
 * $\mathscr A_1 = \set{U \cup \set y : y \in \set{x_1, x_2, \dots, x_k}}$

$\map P 1$ is the case:
 * $\forall j \in \closedint 1 k : \map \rho {X \cup \set{x_j}} = \map \rho U$

Let $j \in \closedint 1 k$.

We have:

By rank axiom $(\text R 2)$:
 * $\map \rho U = \map \rho {U \cup \set {x_j}}$

This establishes our base case.

Induction Hypothesis
Now we need to show that, if $\map P i$ is true, where $i \ge 1$, then it logically follows that $\map P {i + 1}$ is true.

So this is our induction hypothesis:
 * $\forall Y \in \mathscr A_i : \map \rho Y = \map \rho U$

Then we need to show:
 * $\forall Y \in \mathscr A_{i+1} : \map \rho Y = \map \rho U$

Induction Step
This is our induction step:

Let $Y \in \mathscr A_{i+1}$.

Then:
 * $Y = U \cup \set{x_1, x_2, \dots, x_{i+1}, y}$

where $y \in {x_{i+2}, \dots, x_k}$

We note that:
 * $U \cup \set{x_1, x_2, \dots, x_i, x_{i+1}} \in \mathscr A_i$
 * $U \cup \set{x_1, x_2, \dots, x_i, y} \in \mathscr A_i$

We have:

Hence:
 * $\map \rho {U \cup \set{x_1, x_2, \dots, x_i, x_{i+1} } } = \map \rho {U \cup \set{x_1, x_2, \dots, x_i} }$

and
 * $\map \rho {U \cup \set{x_1, x_2, \dots, x_i} } = \map \rho U$

Similarly:
 * $\map \rho {U \cup \set{x_1, x_2, \dots, x_i, y } } = \map \rho {U \cup \set{x_1, x_2, \dots, x_i} }$

By Rank axiom $(\text R 2)$:
 * $\map \rho {U \cup \set{x_1, x_2, \dots, x_i, x_{i+1}, y } } = \map \rho {U \cup \set{x_1, x_2, \dots, x_i} } = \map \rho U$

It follows that:
 * $\forall Y \in \mathscr A_{i+1} : \map \rho Y = \map \rho U$

This establishes our induction step.

Hence by induction:
 * $\forall i \in \closedint 1 k : \forall Y \in \mathscr A_i : \map \rho Y = \map \rho U$

We have:

Hence:
 * $V \notin \mathscr I$