Matroid Unique Circuit Property/Proof 2

This article needs proofreading. Please check it for mathematical errors. If you believe there are none, please remove {{Proofread}} from the code. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Proofread}} from the code.

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

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:

$C \nsubseteq C'$

By the definition of a subset:

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

By the definition of minimal dependent subset:

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

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 $(\text I 6)$:

$\card Y = \card X$

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

$x \ne y$

Hence:

$x \in C \setminus \set y$

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

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

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:

\(\ds \card {\paren {X \cup \set x} \setminus \set y}\)

\(=\)

\(\ds \card {\paren {X \cup \set x} } - \card {\set y}\)

\(\ds \)

\(=\)

\(\ds \card X + \card {\set x} - \card {\set y}\)

\(\ds \)

\(=\)

\(\ds \card X + 1 - 1\)

\(\ds \)

\(=\)

\(\ds \card X\)

\(\ds \)

\(=\)

\(\ds \card Y\)

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:

$C \subseteq X \cup \set x$

$\blacksquare$