# Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Rank Axioms/Condition 1 Implies Condition 3/Lemma 1

## 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:

 $(\text R 1)$ $:$ $\ds \map \rho \O = 0$ $(\text R 2)$ $:$ $\ds \forall X \in \powerset S \land y \in S:$ $\ds \map \rho X \le \map \rho {X \cup \set y} \le \map \rho X + 1$ $(\text R 3)$ $:$ $\ds \forall X \in \powerset S \land y, z \in S:$ $\ds \map \rho {X \cup \set y} = \map \rho {X \cup \set z} = \map \rho X \implies \map \rho {X \cup \set y \cup \set z} = \map \rho X$

Then:

$\forall A, B \subseteq S: A \subseteq B \implies \map \rho A \le \map \rho B$

## Proof

$\exists A, B \subseteq S : A \subseteq B$ and $\map \rho A > \map \rho B$

Let $B \subseteq S$:

$\exists A \subseteq B : \map rho A > \map \rho B$

Let $A_0 \subseteq B$:

$\card {A_0} = \max \set{\card A : A \subseteq B \land \map \rho A > \map \rho B}$

As $\map \rho {A_0} > \map \rho B$:

$A_0 \ne B$
$\exists y \in B \setminus A_0$

We have:

 $\ds \map \rho B$ $<$ $\ds \map \rho {A_0}$ By Choice of $A_0$ $\ds$ $\le$ $\ds \map \rho {A_0 \cup \set y}$ Rank axiom $(\text R 2)$ $\ds$ $\le$ $\ds \map \rho B$ By Choice of $A_0$

$\forall A, B \subseteq S: A \subseteq B \implies \map \rho A \le \map \rho B$
$\blacksquare$