User:Leigh.Samphier/Matroids/Equivalence of Definitions of Matroid Circuit Axioms

From ProofWiki
Jump to navigation Jump to search



Theorem

Let $S$ be a finite set.

Let $\mathscr C$ be a non-empty set of subsets of $S$.


The following definitions for the Matroid Circuit Axioms are equivalent:

Formulation 1

$\mathscr C$ satisfies the circuit axioms:

\((\text C 1)\)   $:$   \(\ds \O \notin \mathscr C \)      
\((\text C 2)\)   $:$     \(\ds \forall C_1, C_2 \in \mathscr C:\) \(\ds C_1 \ne C_2 \implies C_1 \nsubseteq C_2 \)      
\((\text C 3)\)   $:$     \(\ds \forall C_1, C_2 \in \mathscr C:\) \(\ds C_1 \ne C_2 \land z \in C_1 \cap C_2 \implies \exists C_3 \in \mathscr C : C_3 \subseteq \paren {C_1 \cup C_2} \setminus \set z \)      


Formulation 2

$\mathscr C$ satisfies the circuit axioms:

\((\text C 1)\)   $:$   \(\ds \O \notin \mathscr C \)      
\((\text C 2)\)   $:$     \(\ds \forall C_1, C_2 \in \mathscr C:\) \(\ds C_1 \ne C_2 \implies C_1 \nsubseteq C_2 \)      
\((\text C 3')\)   $:$     \(\ds \forall C_1, C_2 \in \mathscr C:\) \(\ds C_1 \ne C_2 \land z \in C_1 \cap C_2 \land w \in C_1 \setminus C_2 \implies \exists C_3 \in \mathscr C : w \in C_3 \subseteq \paren {C_1 \cup C_2} \setminus \set z \)      


Formulation 3

$\mathscr C$ satisfies the circuit axioms:

\((\text C 1)\)   $:$   \(\ds \O \notin \mathscr C \)      
\((\text C 2)\)   $:$     \(\ds \forall C_1, C_2 \in \mathscr C:\) \(\ds C_1 \ne C_2 \implies C_1 \nsubseteq C_2 \)      
\((\text C 3'')\)   $:$     \(\ds \forall X \subseteq S \land \forall x \in S:\) \(\ds \paren {\forall C \in \mathscr C : C \nsubseteq X} \implies \paren {\exists \text{ at most one } C \in \mathscr C : C \subseteq X \cup \set x} \)      


Proof

Formulation 1 implies Formulation 2

Let $\mathscr C$ satisfy the circuit axioms $(\text C 1)$, $(\text C 2)$ and $(\text C 3)$.

It has only to be shown that circuit axiom $(\text C 3')$ is satisfied by $\mathscr C$.


Let:

\(\ds F = \leftset{\tuple{C, D, x, y} }\) \(:\) \(\ds C, D \in \mathscr C \land C \neq D\)
\(\ds \) \(\land\) \(\ds x \in C \cap D \land y \in C \setminus D\)
\(\ds \) \(\land\) \(\ds \nexists C' \in \mathscr C : y \in C' \subseteq \paren{C \cup D} \setminus \set x \mathop {\rightset {} }\)

To show that $\mathscr C$ satisfies circuit axiom $(\text C 3')$, it needs to be shown that $F = \O$.


Aiming for a contradiction, suppose :

$F \neq \O$

Let $\tuple{C_1, C_2, z, w} \in F$ :

$\size{C_1 \cup C_2} = \min \set{\size{C \cup D} : \tuple{C, D, x, y} \in F}$


By circuit axiom $(\text C 3)$:

$\exists C_3 \in \mathscr C : C_3 \subseteq \paren{C_1 \cup C_2} \setminus \set z$

By assumptiom:

$w \notin C_3$


Consider $C_3 \cap \paren{C_2 \setminus C_1}$.

By circuit axiom $(\text C 2)$:

$C_3 \nsubseteq C_1$

From Set Difference and Intersection form Partition:

$C_3 \cap \paren{C_2 \setminus C_1} \neq \O$

Let $x \in C_3 \cap \paren{C_2 \setminus C_1}$.

We have:

$x \in C_3 \cap C_2$

and

$z \in C_2 \setminus C_3$

and

$w \notin C_2 \cup C_3$


From Set is Subset of Union and Union of Subsets is Subset:

$C_2 \cup C_3 \subseteq \C_1 \cup C_2$

Since $w \notin C_2 \cup C_3$:

$C_2 \cup C_3$ is a proper subset of $C_1 \cup C_2$

By circuit axiom $(\text C 3)$ and the minimality of $C_1 \cup C_2$:

$\exists C_4 \in \mathscr C : z \in C_4 \subseteq \paren{C_2 \cup C_3} \setminus \set{x}$


Now consider $C_1$ and $C_4$, we have:

$z \in C_1 \cap C_4$

Since $w \notin C_2 \cup C_3$ then:

$w \in C_1 \setminus C_4$

We have:

$C_4 \subset C_2 \cup C_3 \subset C_1 \cup C_2$

From Set is Subset of Union and Union of Subsets is Subset:

$C_1 \cup C_4 \subseteq C_1 \cup C_2$

Recall $x \in C_3 \cap \paren{C_2 \setminus C_1}$, then:

$x \in C_2$

and

$x \notin C_1$

Since $C_4 \subseteq \paren{C_2 \cup C_3} \setminus \set{x}$, then:

$x \notin C_4$

It follows that:

$x \notin C_1 \cup C_4$

and

$x \in C_1 \cup C_2$

Hence:

$C_1 \cup C_4$ is a proper subset of $C_1 \cup C_2$

By circuit axiom $(\text C 3)$ and the minimality of $C_1 \cup C_2$:

$\exists C_5 \in \mathscr C : w \in C_5 \subseteq \paren{C_1 \cup C_4} \setminus \set{z}$

Since $C_1 \cup C_4 \subset C_1 \cup C_2$, then we have found $C_5$ such that:

$w \in C_5 \subseteq \paren{C_1 \cup C_2} \setminus \set{z}$

This contradicts the fact that $\tuple{C_1, C_2, z, w} \in F$.


It follows that circuit axiom $(\text C 3')$ is satisfied.

$\Box$

Formulation 2 implies Formulation 1

Let $\mathscr C$ satisfy the circuit axioms $(\text C 1)$, $(\text C 2)$ and $(\text C 3')$.

We need to show that $\mathscr C$ satisfies circuit axiom:

\((\text C 3)\)   $:$     \(\ds \forall C_1, C_2 \in \mathscr C:\) \(\ds C_1 \ne C_2 \land z \in C_1 \cap C_2 \implies \exists C_3 \in \mathscr C : C_3 \subseteq \paren {C_1 \cup C_2} \setminus \set z \)      


Let $C_1, C_2 \in \mathscr C : C_1 \ne C_2$.

Let $z \in C_1 \cap C_2$.

From circuit axiom $(\text C 2)$:

$C_2 \nsubseteq C_1$

By definition of subset and set difference:

$\exists w \in C_2 \setminus C_1$

From circuit axiom $(\text C 3')$:

$\exists C_3 \in \mathscr C : w \in C_3 \subseteq \paren{C_1 \cup C_2} \setminus \set z$


It follows that $\mathscr C$ satisfies circuit axiom $(\text C 3)$

$\Box$

Formulation 1 implies Formulation 3

Let $\mathscr C$ satisfy the circuit axioms $(\text C 1)$, $(\text C 2)$ and $(\text C 3)$.

We need to show that $\mathscr C$ satisfies circuit axiom:

\((\text C 3'')\)   $:$     \(\ds \forall X \subseteq S \land \forall x \in S:\) \(\ds \paren {\forall C \in \mathscr C : C \nsubseteq X} \implies \paren {\exists \text{ at most one } C \in \mathscr C : C \subseteq X \cup \set x} \)      


Let $X \subset S : \forall C \in \mathscr C : C \nsubseteq X$.

Let $x \in S$.


Aiming for a contradiction, suppose:

$\exists C_1, C_2 \in \mathscr C : C_1 \neq C_2 : C_1, C_2 \subseteq X \cup \set x$.

Since $C_1, C_2 \nsubseteq X$ then:

$x \in C_1 \cap C_2$

From circuit axiom $(\text C 3)$:

$\exists C_3 \in \mathscr C : C_3 \subseteq \paren{C_1 \cup C_2} \setminus \set x$

We have:

$\paren{C_1 \cup C_2} \setminus \set x \subseteq X$

Hence:

$C_3 \subseteq X$

This contradicts the assumption that:

$X \subset S : \forall C \in \mathscr C : C \nsubseteq X$


It follows that:

$\exists \text{ at most one } C \in \mathscr C : C \subseteq X \cup \set x$

It follows that $\mathscr C$ satisfies circuit axiom $(\text C 3'')$.

$\Box$

Formulation 3 implies Formulation 1

Let $\mathscr C$ satisfy the circuit axioms $(\text C 1)$, $(\text C 2)$ and $(\text C 3'')$.

We need to show that $\mathscr C$ satisfies circuit axiom:

\((\text C 3)\)   $:$     \(\ds \forall C_1, C_2 \in \mathscr C:\) \(\ds C_1 \ne C_2 \land z \in C_1 \cap C_2 \implies \exists C_3 \in \mathscr C : C_3 \subseteq \paren {C_1 \cup C_2} \setminus \set z \)      

In fact we prove the contrapositive statement:

  \(\ds \forall C_1, C_2 \in \mathscr C:\) \(\ds z \in C_1 \cap C_2 \land \paren{\forall C \in \mathscr C : C \nsubseteq \paren {C_1 \cup C_2} \setminus \set z} \implies C1 = C2 \)      


Let:

$C_1, C_2 \in \mathscr C$
$z \in C_1 \cap C_2$
$\forall C \in \mathscr C : C \nsubseteq \paren{C_1 \cup C_2} \setminus \set z$


From circuit axiom $(\text C 3'')$:

$\exists \text{ at most one } C \in \mathscr C : C \subseteq \paren{\paren{C_1 \cup C_2} \setminus \set z} \cup \set z = C_1 \cup C_2$

From Set is Subset of Union:

$C_1, C_2 \subseteq C_1 \cup C_2$

Hence:

$C_1 = C_2$

It follows that $\mathscr C$ satisfies circuit axiom $(\text C 3)$.

$\blacksquare$


Also see