# Matroid Unique Circuit Property

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## 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:

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

### Corollary

Let $B$ be a base of $M$.

Let $x \in S \setminus B$.

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

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

That is, $C$ is the fundamental circuit of $x$ in $B$.

## Proof 1

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.

$\blacksquare$

## Proof 2

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:

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

By the definition of minimal dependent subset:

$C \nsubseteq C'$

By the definition of a subset:

$\exists y \in C \setminus C'$

By the definition of minimal dependent subset:

$C \setminus \set y \in \mathscr I$
$\exists Y \in \mathscr I : C \setminus \set y \subseteq Y \subseteq X \cup \set x : Y$ is a maximal independent subset of $X \cup \set x$

By assumption:

$X$ is a maximal independent subset of $X \cup \set x$
$\card Y = \card X$

As $x \in C'$ and $y \notin C'$ then:

$x \ne y$

Hence:

$x \in C \setminus \set y$
$C \setminus \set y \cup \set y = C \nsubseteq Y$

Hence:

$y \notin Y$

Hence:

$Y \subseteq \paren {X \cup \set x} \setminus \set y$
$\card Y \le \card {\paren {X \cup \set x} \setminus \set y}$

We have:

 $\ds \card {\paren {X \cup \set x} \setminus \set y}$ $=$ $\ds \card {\paren {X \cup \set x} } - \card {\set y}$ $\ds$ $=$ $\ds \card X + \card {\set x} - \card {\set y}$ $\ds$ $=$ $\ds \card X + 1 - 1$ $\ds$ $=$ $\ds \card X$ $\ds$ $=$ $\ds \card Y$
$Y = \paren {X \cup \set x} \setminus \set y$

Hence:

$\paren {X \cup \set x} \setminus \set y \in \mathscr I$

As $y \notin C'$ and $C' \subseteq X \cup \set x$ then:

$C' \subseteq \paren {X \cup \set x} \setminus \set y$
$\paren {X \cup \set x} \setminus \set y \notin \mathscr I$

This contradicts the previous statement that

$\paren {X \cup \set x} \setminus \set y \in \mathscr I$

It follows that $C$ is the unique circuit such that:

$C \subseteq X \cup \set x$

$\blacksquare$