User:Leigh.Samphier/Matroids/Equivalence of Definitions of Matroid Circuit Axioms/Formulation 1 Implies Formulation 3

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 (formulation 1):

Then:
 * $\mathscr C$ satisfies the circuit axioms (formulation 3):

Proof
Let $\mathscr C$ satisfy the circuit axioms $(\text C 1)$, $(\text C 2)$ and $(\text C 3)$.

We need to show that $\mathscr C$ satisfies circuit axiom:

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

Let $x \in S$.


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

Since $C_1, C_2 \nsubseteq X$ then:
 * $x \in C_1 \cap C_2$

From $(\text C 3)$:
 * $\exists C_3 \in \mathscr C : C_3 \subseteq \paren{C_1 \cup C_2} \setminus \set x$

We have:
 * $\paren{C_1 \cup C_2} \setminus \set x \subseteq X$

Hence:
 * $C_3 \subseteq X$

This contradicts the assumption that:
 * $X \subset S : \forall C \in \mathscr C : C \nsubseteq X$

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 ${\text C 3''}$.