# Equivalence of Definitions of Matroid Circuit Axioms

## Theorem

Let $S$ be a finite set.

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

The following statements are equivalent:

### Condition 1

$\mathscr C$ satisfies the circuit axioms:

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

### Condition 2

$\mathscr C$ satisfies the circuit axioms:

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

### Condition 3

$\mathscr C$ satisfies the circuit axioms:

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

### Condition 4

$\mathscr C$ is the set of circuits of a matroid on $S$

## Proof

### Condition 1 implies Condition 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 $\tuple{C_1, C_2, z, w} \in F$ :

$\size{C_1 \cup C_2} = \min \set{\size{C \cup D} : \tuple{C, D, x, y} \in F}$
$\exists C_3 \in \mathscr C : C_3 \subseteq \paren{C_1 \cup C_2} \setminus \set z$

By assumptiom:

$w \notin C_3$

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

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

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

$w \notin C_2 \cup C_3$
$C_2 \cup C_3 \subseteq \C_1 \cup C_2$

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:

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

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.

$\Box$

### Condition 2 implies Condition 4

We will define a mapping $\rho$ associated with $\mathscr C$.

It will be shown that $\rho$ is the rank function of a matroid $M$ which has $\mathscr C$ as the set of circuits.

For any ordered tuple $\tuple{x_1, \ldots, x_q}$ of elements of $S$, let $\map \theta {x_1, \ldots, x_q}$ be the ordered tuple defined by:

$\forall i \in \set{1, \ldots, q} : \map \theta {x_1, \ldots, x_q}_i = \begin{cases} 0 & : \exists C \in \mathscr C : x_i \in C \subseteq \set{x_1, \ldots, x_i}\\ 1 & : \text {otherwise} \end{cases}$

Let $t$ be the mapping from the set of ordered tuple of $S$ defined by:

$\map t {x_1, \ldots, x_q} = \ds \sum_{i = 1}^q \map \theta {x_1, \ldots, x_q}_i$

#### Lemma 1

Let $\tuple{x_1, \ldots, x_q}$ be any ordered tuple of elements of $S$.

Let $\pi$ be any permutation of $\tuple{x_1, \ldots, x_q}$.

Then:

$\map t {x_1, \ldots, x_q} = \map t {x_{\map \pi 1}, \ldots, x_{\map \pi q}}$

$\Box$

Let $\rho : \powerset S \to \Z$ be the mapping defined by:

$\forall A \subseteq S$:
$\map \rho A = \begin{cases} 0 & : \text{if } A = \O \\ \map t {x_1, \ldots, x_q } & : \text{if } A = \set{x_1, \ldots, x_q} \end{cases}$

From Lemma 1:

$\rho$ is well-defined

We now show that $\rho$ satisfies the rank function axioms:

 $(\text R 1)$ $:$ $\ds \map \rho \O = 0$ $(\text R 2)$ $:$ $\ds \forall X \in \powerset S \land y \in S:$ $\ds \map \rho X \le \map \rho {X \cup \set y} \le \map \rho X + 1$ $(\text R 3)$ $:$ $\ds \forall X \in \powerset S \land y, z \in S:$ $\ds \map \rho {X \cup \set y} = \map \rho {X \cup \set z} = \map \rho X \implies \map \rho {X \cup \set y \cup \set z} = \map \rho X$

#### $\rho$ satisfies $(\text R 1)$

Follows from the definition of $\rho$.

$\Box$

#### $\rho$ satisfies $(\text R 2)$

Let:

$X = \set{x_1, \ldots, x_q} \subseteq S$ and $y \in S$

We have:

 $\ds \map \rho {X \cup \set y} - \map \rho X$ $=$ $\ds \map t {\tuple{x_1, \dots, x_q, y} } - \map t {\tuple{x_1, \dots, x_q} }$ Definition of $\rho$ $\ds$ $=$ $\ds \map \theta {\tuple{x_1, \dots, x_q, y} }_{q+1}$ Definition of $t$ $\ds$ $=$ $\ds \begin{cases} 0 & : \exists C \in \mathscr C : y \in C \subseteq \set{x_1, \ldots, x_q, y}\\ 1 & : \text {otherwise} \end{cases}$ Definition of $\theta$ $\ds \leadstoandfrom \ \$ $\ds 0$ $\le$ $\ds \map \rho {X \cup \set y} - \map \rho X \le 1$ $\ds \leadstoandfrom \ \$ $\ds \map \rho X$ $\le$ $\ds \map \rho {X \cup \set y} \le \map \rho X + 1$ Adding $\map \rho X$ to all terms

$\Box$

#### Lemma 2

Let $X \subseteq S$ and $y \in S \setminus X$.

Then:

$\map \rho {X \cup \set y} = \map \rho X$ if and only if $\exists C \in \mathscr C : y \in C \subseteq X \cup \set y$

$\Box$

#### $\rho$ satisfies $(\text R 3)$

Let:

$X \subseteq S$ and $y, z \in S$

Let:

$\map \rho {X \cup \set y} = \map \rho {X \cup \set z} = \map \rho X$
##### Case 1 : $z \in X$

Let $z \in X$.

Then:

$\map \rho {X \cup \set {y,z}} = \map \rho {X \cup \set y}$

Since $\map \rho {X \cup \set y} = \map \rho X$, then:

$\map \rho {X \cup \set {y,z}} = \map \rho X$

$\Box$

##### Case 2 : $z = y$

Let $z = y$.

Then

$\map \rho {X \cup \set {y,z}} = \map \rho {X \cup \set y}$

Since $\map \rho {X \cup \set y} = \map \rho X$, then:

$\map \rho {X \cup \set {y,z}} = \map \rho X$

$\Box$

##### Case 3 : $z \neq y$ and $z \notin X$

Let $z \neq y$ and $z \notin X$.

Then $z \notin X \cup \set y$.

From Lemma 2:

$\exists C_z \in \mathscr C : z \in C_z \subseteq X \cup \set{z} \subseteq X \cup \set{y,z}$

From Lemma 2:

$\map \rho {X \cup \set{y, z}} = \map \rho {X \cup \set y} = \map \rho X$

$\Box$

$\rho$ is the rank function of a matroid $M = \struct{S, \mathscr I}$ on $S$

#### Lemma 3

$\mathscr C$ is the set of circuits of $M$.

$\Box$

### Condition 4 implies Condition 1

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

#### $\mathscr C$ 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 $(\text C 3)$.

$\Box$

### Condition 1 implies Condition 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 $x \in S$.

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

Since $C_1, C_2 \nsubseteq X$ then:

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

We have:

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

Hence:

$C_3 \subseteq X$

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

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 $(\text C 3'')$.

$\Box$

### Condition 3 implies Condition 1

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:

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

Hence:

$C_1 = C_2$

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

$\blacksquare$