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

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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:

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


Then:

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


Proof

Aiming for a contradiction, suppose:

$\exists A, B \subseteq S : A \subseteq B$ 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 $\map \rho {A_0} > \map \rho B$:

$A_0 \ne B$

From Set Difference with Proper Subset:

$\exists y \in B \setminus A_0$


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 $A_0$

This is a contradiction.


Hence:

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

$\blacksquare$