Equivalence of Definitions of Matroid Circuit Axioms/Condition 4 Implies Condition 1

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 $\mathscr C$ be the set of circuits of a matroid $M = \struct{S, \mathscr I}$ on $S$


Then:

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


Proof

$\mathscr C$ satisfies $(\text C 1)$

By definition of circuit of a matroid:

for all $C \in \mathscr C: C$ is a dependent subset

By definition of a dependent subset:

for all $C \in \mathscr C$, $C \notin \mathscr I$

By definition of a matroid:

$\O \in \mathscr I$

Hence:

$\O \notin \mathscr C$

It follows that $\mathscr C$ satisfies circuit axiom $(\text C 1)$.

$\Box$

$\mathscr C$ satisfies $(\text C 2)$

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

By definition of circuit of a matroid:

$C_2$ is a dependent subset of $S$ which is a minimal dependent subset

and

$C_1$ is a dependent subset

Hence:

$C_1 \nsubseteq C_2$

It follows that $\mathscr C$ satisfies circuit axiom $(\text C 2)$.

$\Box$

$\mathscr C$ satisfies $(\text C 3)$

Let $\rho$ denote the rank function of the matroid $M$.


Aiming for a contradiction, suppose $C_1, C_2 \in \mathscr C:$

$C_1 \neq C_2$

and

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


From Proper Subset of Matroid Circuit is Independent:

$\paren{C_1 \cup C_2} \setminus \set z$ is an independent subset.

Similarly:

$\paren{C_1 \cap C_2}$ is an independent subset.


We have:

\(\ds \map \rho {C_1 \cup C_2}\) \(\le\) \(\ds \map \rho {C_1} + \map \rho {C_2} - \map \rho {C_1 \cap C_2}\) Rank axiom $(\text R 3')$
\(\ds \) \(=\) \(\ds \paren{\size {C_1} - 1} + \paren{\size {C_2} - 1} - \map \rho {C_1 \cap C_2}\) Rank of Matroid Circuit is One Less Than Cardinality
\(\ds \) \(=\) \(\ds \size {C_1} + \size {C_2} - 2 - \map \rho {C_1 \cap C_2}\) Rank of Matroid Circuit is One Less Than Cardinality
\(\ds \) \(=\) \(\ds \size {C_1 \cup C_2} + \size {C_1 \cap C_2} - 2 - \map \rho {C_1 \cap C_2}\) Cardinality of Set Union
\(\ds \) \(=\) \(\ds \size {C_1 \cup C_2} + \size {C_1 \cap C_2} - 2 - \size {C_1 \cap C_2}\) Rank of Independent Subset Equals Cardinality
\(\ds \) \(=\) \(\ds \size {C_1 \cup C_2} - 2\) Cancelling terms
\(\ds \) \(=\) \(\ds \size {C_1 \cup C_2} - \size { \set z} - 1\) Cardinality of Singleton
\(\ds \) \(=\) \(\ds \size {\paren{C_1 \cup C_2} \setminus \set z} - 1\) Cardinality of Set Difference with Subset
\(\ds \) \(=\) \(\ds \map \rho {\paren{C_1 \cup C_2} \setminus \set z} - 1\) Rank of Independent Subset Equals Cardinality
\(\ds \) \(\le\) \(\ds \map \rho {\paren{C_1 \cup C_2} } - 1\) Rank Function is Increasing

This is a contradiction.

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

$\blacksquare$

Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Equivalence_of_Definitions_of_Matroid_Circuit_Axioms/Condition_4_Implies_Condition_1&oldid=613188"