Formulation 1 Rank Axioms Implies Rank Function of Matroid/Lemma 5/Proof 2

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Theorem

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:

$\forall x \in V \setminus U: U \cup \set x \notin \mathscr I$

That is, by definition of $\mathscr I$:

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

We have:

\(\ds \card V\)

\(>\)

\(\ds \card U\)

by hypothesis

\(\ds \)

\(=\)

\(\ds \map \rho U\)

As $U \in \mathscr I$

\(\ds \)

\(=\)

\(\ds \map \rho {U \cup V}\)

Lemma 3

\(\ds \)

\(\ge\)

\(\ds \map \rho V\)

Lemma 1

Hence:

$V \notin \mathscr I$

It follows:

$M$ satisfies matroid axiom $(\text I 3)$.

Lemma 2

$\forall A \subseteq S: \map \rho A \le \card A$

Proof

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 $X \notin \mathscr I$.

$\Box$

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:

\(\ds \map \rho X\)

\(=\)

\(\ds \map \rho {Y \cup \set{z_1, z_2, \ldots, z_k} }\)

\(\ds \)

\(\le\)

\(\ds \map \rho Y + k\)

Repeated application of rank axiom $(\text R 2)$

\(\ds \)

\(<\)

\(\ds \size Y + k\)

Lemma 2

\(\ds \)

\(=\)

\(\ds \size Y + \size {\set {z_1, z_2, \ldots, z_k} }\)

Definition of Cardinality of Finite Set

\(\ds \)

\(=\)

\(\ds \size {Y \cup \set {z_1, z_2, \ldots, z_k} }\)

Corollary of Cardinality of Set Union

\(\ds \)

\(=\)

\(\ds \size X\)

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

$\Box$

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 $(\text I 2)$.

$\blacksquare$