Equivalence of Definitions of Matroid Circuit Axioms/Condition 1 Implies Condition 2

From ProofWiki
Jump to navigation Jump to search


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


Let $\mathscr C$ satisfy the circuit axioms:

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


Then:

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


Proof

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.

$\blacksquare$

Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Equivalence_of_Definitions_of_Matroid_Circuit_Axioms/Condition_1_Implies_Condition_2&oldid=613186"