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

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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:

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


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


Proof

Lemma 1
$\forall A, B \subseteq S: A \subseteq B \implies \map \rho A \le \map \rho B$

$\Box$


Lemma 2
$\forall A \subseteq S: \map \rho A \le \card A$

$\Box$


Lemma 3

Let:

$A \subseteq S : \map \rho A = \card A$

Let:

$B \subseteq S : \forall b \in B \setminus A : \map \rho {A \cup \set b} \ne \card{A \cup \set b}$


Then:

$\map \rho {A \cup B} = \map \rho A$

$\Box$


Let:

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

It is to be shown that:

and


Matroid Axioms

Matroid Axiom $(\text I 1)$

We have:

\(\ds \map \rho \O\) \(=\) \(\ds 0\) Rank axiom $(\text R 1)$
\(\ds \) \(=\) \(\ds \card \O\) Cardinality of Empty Set

So:

$\O \in \mathscr I$

Hence:

$M$ satisfies matroid axiom $(\text I 1)$.

$\Box$


Matroid Axiom $(\text I 2)$

Let

$X \in \mathscr I$


Aiming for a contradiction, suppose

$\exists Y \subseteq X : Y \notin \mathscr I$

Let:

$Y_0 \subseteq X : \card {Y_0} = \max \set{\card Z : Z \subseteq X \land Z \notin \mathscr I}$


By definition of $\mathscr I$:

$Y_0 \notin \mathscr I \leadsto \map \rho {Y_0} \ne \card {Y_0}$

From Lemma 2:

$\map \rho {Y_0} < \card {Y_0}$


As $X \in \mathscr I$ then:

$Y_0 \ne X$

From Set Difference with Proper Subset:

$X \setminus Y_0 \ne \O$


Let $y \in X \setminus Y_0$.


We have:

\(\ds \card {Y_0} + 1\) \(=\) \(\ds \card {Y_0 \cup \set y}\) Corollary of Cardinality of Set Union
\(\ds \) \(=\) \(\ds \map \rho {Y_0 \cup \set y}\) As $\card {Y_0} = \max \set{\card Z : Z \subseteq X \land Z \notin \mathscr I}$
\(\ds \) \(\le\) \(\ds \map \rho {Y_0} + 1\) Rank axiom $(\text R 2)$
\(\ds \) \(<\) \(\ds \card {Y_0} + 1\) As $\map \rho {Y_0} < \card {Y_0}$

This is a contradiction.


So:

$\forall Y \subseteq X : Y \in \mathscr I$

Hence:

$M$ satisfies matroid axiom $(\text I 2)$.

$\Box$


Matroid Axiom $(\text I 3)$

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

That is:

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


We have:

\(\ds \card V\) \(>\) \(\ds \card U\)
\(\ds \) \(=\) \(\ds \map \rho U\) As $U \in \mathscr I$
\(\ds \) \(=\) \(\ds \map \rho {U \cup V}\) Lemma 3
\(\ds \) \(\ge\) \(\ds \map \rho V\) Lemma 1

Hence:

$V \notin \mathscr I$

$\Box$


$\rho$ is Rank Function

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

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 $M = \struct{S, \mathscr I}$.

$\blacksquare$


Sources