Equivalence of Definitions of Matroid Rank Axioms/Condition 1 Implies Condition 3/Proof 2

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 formulation 1 of the rank axioms:

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

Lemma 2
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:

Hence $\O \in \mathscr I$, and $\mathscr I$ satisfies matroid Axiom $(\text I 1)$.

Matroid Axiom $(\text I 2)$
We prove the contrapositive statement:
 * $\forall X, Y \subseteq S: Y \notin \mathscr I \land Y \subseteq X \implies X \notin \mathscr I$

Let $X, Y \subseteq S : Y \notin \mathscr I$ and $Y \subseteq X$.

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

Then $X \notin \mathscr I$.

Case 2: $Y \subset X$
Let $Y \subset X$.

By definition of $\mathscr I$:
 * $\map \rho Y \neq \size Y$

From Lemma 2:
 * $\map \rho Y < \size Y$

Let:
 * $X \setminus Y = \set{z_1, z_2, \ldots, z_k}$

We have:

By repeated application of rank axiom $(\text R 2)$, we have:

It follows that $X \notin \mathscr I$.

In either case we have:
 * $X \notin \mathscr I$

This proves the contrapositive statement:
 * $\forall X, Y \subseteq S: Y \notin \mathscr I \land Y \subseteq X \implies X \notin \mathscr I$

It follows that $\mathscr I$ satisfies matroid Axiom $(\text I 2)$.

Matroid Axiom $(\text I 3)$
It is now proved that $\mathscr I$ satisifes the matroid Axiom $(\text I 3')$:

Let $X, Y \in \mathscr I$ such that $\size Y = \size X + 1$.

Let:
 * $X = \set {x_1, x_2, \ldots, x_q, z_{q + 1}, \ldots, z_k}$

and
 * $Y = \set {x_1, x_2, \ldots, x_q, y_{q + 1}, \ldots, y_k, y_{k + 1}}$

where $\forall i \in \set {q + 1, \ldots, k}$ and $\forall j \in \set {q + 1, \ldots, k + 1}$
 * $z_i \neq y_j$


 * $\forall j \in \set {q + 1, \ldots, k + 1}: X \cup \set{y_j} \notin \mathscr I$
 * $\forall j \in \set {q + 1, \ldots, k + 1}: X \cup \set{y_j} \notin \mathscr I$

We have:

Hence:
 * $\size X \le \map \rho {X \cup \set{y_j} } < \size {X \cup \set{y_j} } = \size X + 1$

It follows that:
 * $\size X = \map \rho {X \cup \set{y_j} }$

Hence:
 * $\forall j \in \set {q + 1, \ldots, k + 1}: \map \rho X = \map \rho {X \cup \set{y_j} } = \size X$

By rank axiom $(\text R 3)$:
 * $\forall i, j \in \set {q + 1, \ldots, k + 1}: \map \rho {X \cup \set{y_i, y_j} } = \map \rho X = \size X$

Applying rank axiom $(\text R 3)$ repeatedly, we get:
 * $\map \rho {X \cup \set{y_{q + 1}, \ldots, y_{k + 1} } = \map \rho X = \size X < \size Y$

Now:
 * $Y \subseteq X \cup \set{y_{q + 1}, \ldots, y_{k + 1} }$

From Lemma 1:
 * $\map \rho Y \le \map \rho {X \cup \set{y_{q + 1}, \ldots, y_{k + 1} } < \size Y$

This contrdict the assumption that $Y \in \mathscr I$.

Hence:
 * $\exists j \in \set {q + 1, \ldots, k + 1}: X \cup \set{y_j} \in \mathscr I$

It follows that $\mathscr I$ satisfies matroid Axiom $(\text I 3')$.

We have proven that $\mathscr I$ satisifies the matroid axioms.

It follows that $M = \struct{S, \mathscr I}$ is a matroid by definition.

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