Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Rank Axioms

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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.


The following are equivalent:

Condition 1

$\rho$ satisfies definition 1 of the rank axioms:

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


Condition 2

$\rho$ satisfies definition 2 of 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 $M = \struct{S, \mathscr I}$.


Proof

Condition 1 implies Condition 3

Let $\rho$ satisfy definition 1 of the rank axioms:

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


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:

\(\displaystyle \map \rho \O\) \(=\) \(\displaystyle 0\) Rank axiom $(\text R 1)$
\(\displaystyle \) \(=\) \(\displaystyle \card \O\) 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:

$X \setminus Y_0 \neq \O$


Let $y \in X \setminus Y_0$.


We have:

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

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

\(\displaystyle \map {\rho_M} X\) \(=\) \(\displaystyle \card {Y_0}\) By choice of $Y_0$
\(\displaystyle \) \(=\) \(\displaystyle \map \rho {Y_0}\) 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 $M = \struct{S, \mathscr I}$.

$\Box$


Condition 3 implies Condition 2

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


$\rho$ satisfies $(\text R 1')$

This follows immediately from Bounds for Rank of Subset.

$\Box$


$\rho$ satisfies $(\text R 2')$

This follows immediately from Rank Function is Increasing.

$\Box$


$\rho$ satisfies $(\text R 3')$

Let $X, Y \subseteq S$.


Let $A$ be a maximal independent subset of $X \cap Y$.

From Leigh.Samphier/Sandbox/Independent Subset is Contained in Maximal Independent Subset/Corollary:

$\card A = \map \rho {X \cap Y}$


From Independent Subset is Contained in Maximal Independent Subset:

$\exists B \subseteq X \setminus Y : A \cup B$ is a maximal independent subset of $X$

Simiarly from Independent Subset is Contained in Maximal Independent Subset:

$\exists C \subseteq Y \setminus X : \paren{A \cup B} \cup C$ is a maximal independent subset of $X \cup Y$

and

$\card{A \cup B \cup C} = \map \rho {X \cup Y}$


By matroid axiom $(\text I 2)$:

$A \cup C$ is an independent subset of $Y$.


We have:

\(\displaystyle \map \rho X + \map \rho Y\) \(\ge\) \(\displaystyle \card{A \cup B} + \card{A \cup C}\) Definition of Rank function $\rho$
\(\displaystyle \) \(=\) \(\displaystyle \card{A \cup B} + \card A + \card C\) Corollary to Cardinality of Set
\(\displaystyle \) \(=\) \(\displaystyle \card{A \cup B \cup C} + \card A\) Corollary to Cardinality of Set
\(\displaystyle \) \(=\) \(\displaystyle \map \rho {X \cup Y} + \map \rho {X \cap Y}\) Definition of Rank function $\rho$

$\Box$


Condition 2 implies Condition 1

Let $\rho$ satisfy definition 2 of the rank axioms:

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



$\rho$ satisfies $(\text R 1)$

We have:

\(\displaystyle 0\) \(\le\) \(\displaystyle \map \rho \O\) rank axiom $(\text R 1')$
\(\displaystyle \) \(\le\) \(\displaystyle \card \O\) rank axiom $(\text R 1')$
\(\displaystyle \) \(=\) \(\displaystyle 0\) Cardinality of Empty Set

Hence:

$\map \rho \O = 0$

$\Box$


$\rho$ satisfies $(\text R 2)$

Let $X \subseteq S$.

Let $y \in S$.


We have:

\(\displaystyle \map \rho X\) \(\le\) \(\displaystyle \map \rho {X \cup y}\) rank axiom $(\text R 2')$
\(\displaystyle \) \(\le\) \(\displaystyle \map \rho X + \map \rho {\set y} - \map \rho {X \cup \set y}\) rank axiom $(\text R 3')$
\(\displaystyle \) \(\le\) \(\displaystyle \map \rho X + \map \rho {\set y}\)
\(\displaystyle \) \(\le\) \(\displaystyle \map \rho X + \card {\set y}\) rank axiom $(\text R 1')$
\(\displaystyle \) \(\le\) \(\displaystyle \map \rho X + 1\) Cardinality of Singleton

This proves rank axiom $(\text R 2)$

$\Box$


$\rho$ satisfies $(\text R 3)$

$\blacksquare$


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