User:Leigh.Samphier/Matroids/Circuits of Matroid iff Matroid Circuit Axioms/Circuits of Matroid implies Formulation 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:

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

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

$\mathscr C$ satisfies $(\text C 3)$
Let $\rho$ denote the rank function of the matroid $M$.

$C_1, C_2 \in \mathscr C:$
 * $C_1 \neq C_2$

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:

This is a contradiction.

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