# Leigh.Samphier/Sandbox/Matroid satisfies Rank Axioms

## 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.

The following are equivalent:

### Condition 1

$\rho$ satisfies the rank axioms:

 $(R1)$ $:$ $\displaystyle \map \rho \O = 0$ $(R2)$ $:$ $\displaystyle \forall X \in \powerset S \land y \in S:$ $\displaystyle \map \rho X \le \map \rho {X \cup \set y} \le \map \rho X + 1$ $(R3)$ $:$ $\displaystyle \forall X \in \powerset S \land y, z \in S:$ $\displaystyle \map \rho {X \cup \set y} = \map \rho {X \cup \set z} = \map \rho X \implies \map \rho {X \cup \set{y,z} } = \map \rho X$

### Condition 2

$\rho$ satisfies the rank axioms:

 $(\text R 1')$ $:$ $\displaystyle \forall X \in \powerset S:$ $\displaystyle 0 \le \map \rho X \le \size X$ $(\text R 2')$ $:$ $\displaystyle \forall X, Y \in \powerset S:$ $\displaystyle X \subseteq Y \implies \map \rho X \le \map \rho Y$ $(\text R 3')$ $:$ $\displaystyle \forall X, Y \in \powerset S:$ $\displaystyle \map \rho {X \cup Y} + \map \rho {X \cap Y} \le \map \rho X + \map \rho Y$

### Condition 3

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

## Proof

$\blacksquare$