Equivalence of Definitions of Matroid Rank Axioms/Condition 1 Implies Condition 3/Proof 1
This article needs proofreading. Please check it for mathematical errors. If you believe there are none, please remove {{Proofread}} from the code.To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Proofread}} from the code. |
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:
- $\quad \mathscr I$ satisfies the matroid axioms
and
- $\quad \rho$ is the rank function of the matroid $M = \struct{S, \mathscr I}$
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 $M = \struct{S, \mathscr I}$.
$\blacksquare$