# Leigh.Samphier/Sandbox/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:

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

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:

$C \subseteq X \cup \set x$

$\blacksquare$