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

## 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

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

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