Equivalence of Definitions of Matroid Circuit Axioms/Condition 1 Implies Condition 3

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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:

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

Then:

$\mathscr C$ satisfies the circuit axioms:

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

Proof

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 $x \in S$.

Aiming for a contradiction, suppose:

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

Since $C_1, C_2 \nsubseteq X$ then:

$x \in C_1 \cap C_2$

From circuit axiom $(\text C 3)$:

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

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 $(\text C 3'')$.

$\blacksquare$