# Matroid Unique Circuit Property/Proof 1

## Theorem

Let $M = \struct {S, \mathscr I}$ be a matroid.

Let $X \subseteq S$ be an independent subset of $M$.

Let $x \in S$ such that:

$X \cup \set x$ is a dependent subset of $M$.

Then there exists a unique circuit $C$ such that:

$x \in C \subseteq X \cup \set x$

## Proof

there exists a circuit $C$ such that $C \subseteq X \cup \set x$
$x \in C$

Aiming for a contradiction, suppose $C'$ is circuit of $M$ such that:

$C' \ne C$
$C' \subseteq X \cup \set x$
$x \in C'$

Hence:

$x \in C \cap C'$

From Equivalence of Definitions of Matroid Circuit Axioms, the set $\mathscr C$ of all circuits satisfies the matroid circuit axiom $(\text C 3)$:

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

Hence there exists a circuit $C_3$ of $M$:

$C_3 \subseteq \paren{C \cup C'} \setminus \set x \subseteq X$

This contradicts the independence of $X$.

The result follows.

$\blacksquare$