Equivalence of Definitions of Matroid Rank Axioms/Lemma 4

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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 definition 1 of the rank axioms:

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

Let $M = \struct{S, \mathscr I}$ be the matroid where:

$\mathscr I = \set{X \subseteq S : \map \rho X = \card X}$

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

Then $\rho_M = \rho$.

Proof

Let $X \subseteq S$.

By definition of the rank function:

$\map {\rho_M} X = \max \set{\card Y : Y \subseteq X, Y \in \mathscr I}$

Let $Y_0 \subseteq X$:

$\card {Y_0} = \max \set{\card Y : Y \subseteq X, Y \in \mathscr I}$

We have:

\(\ds \map {\rho_M} X\)

\(=\)

\(\ds \card {Y_0}\)

By choice of $Y_0$

\(\ds \)

\(=\)

\(\ds \map \rho {Y_0}\)

As $Y_0 \in \mathscr I$

So it remains to show:

$\map \rho {Y_0} = \map \rho X$

Case 1 : $Y_0 = X$

Let $Y_0 = X$.

Then:

$\map \rho {Y_0} = \map \rho X$

$\Box$

Case 2 : $Y_0 \ne X$

Let $Y_0 \ne X$.

Then:

$Y_0 \subsetneq X$

From Set Difference with Proper Subset:

$X \setminus Y_0 \ne \O$

By choice of $Y_0$:

$\forall y \in X \setminus Y_0 : Y_0 \cup \set y \notin \mathscr I$

That is:

$\forall y \in X \setminus Y_0 : \map \rho {Y_0 \cup \set y} \ne \card {Y_0 \cup \set y}$

From Lemma 3:

$\map \rho {Y_0} = \map \rho {Y_0 \cup X} = \map \rho X$

$\Box$

In either case:

$\map \rho {Y_0} = \map \rho X$

It follows that:

$\forall X \subseteq S : \map {\rho_M} X = \map \rho X$

Hence $\rho$ is the rank function of the matroid $M = \struct{S, \mathscr I}$.

$\blacksquare$