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

From Dependent Subset Contains a Circuit:

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

From Dependent Subset of Independent Set Union Singleton Contains Singleton:

$x \in C$


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

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


From Dependent Subset of Independent Set Union Singleton Contains Singleton:

$x \in C'$


By the definition of minimal dependent subset:

$C \not \subseteq C'$

By the definition of a subset:

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

By the definition of minimal dependent subset:

$C \setminus \set y \in \mathscr I$


From Independent Subset is Contained in Maximal Independent Subset:

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

By assumption:

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

By matroid axiom $(\text I 3''')$:

$\card Y = \card X$


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

$x \ne y$

Hence:

$x \in C \setminus \set y$


By matroid axiom $(\text I 2)$:

$C \setminus \set y \cup \set y = C \not \subseteq Y$

Hence:

$y \notin Y$

Hence:

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


From Cardinality of Subset of Finite Set:

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

We have:

\(\displaystyle \card{\paren{X \cup \set x} \setminus \set y}\) \(=\) \(\displaystyle \card{\paren{X \cup \set x} } - \card{\set y}\)
\(\displaystyle \) \(=\) \(\displaystyle \card X + \card{\set x} - \card{\set y}\)
\(\displaystyle \) \(=\) \(\displaystyle \card X + 1 - 1\)
\(\displaystyle \) \(=\) \(\displaystyle \card X\)
\(\displaystyle \) \(=\) \(\displaystyle \card Y\)

From Cardinality of Proper Subset of Finite Set:

$Y = \paren{X \cup \set x} \setminus \set y$

Hence:

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


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

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

From Superset of Dependent Set is Dependent:

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

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:

$C \subseteq X \cup \set x$

$\blacksquare$