# Leigh.Samphier/Sandbox/Rank of Matroid Circuit is One Less Than Cardinality/Lemma

## Theorem

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

Let $C \subseteq S$ be a circuit of $M$.

Let $x \in C$.

Then:

$C \setminus \set x$ is a maximal independent subset of $C$

## Proof

$C \setminus \set x \subseteq C$

Because $x \in C$ and $x \notin C \setminus \set x$:

$C \setminus \set x \ne C$
$C \setminus \set x \in \mathscr I$

Let $X$ be an independent subset such that:

$C \setminus \set x \subseteq X \subseteq C$

As $C$ is dependent:

$X \ne C$

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

We have:

 $\ds X$ $=$ $\ds X \cup \set x$ Union with Superset is Superset $\ds$ $\supseteq$ $\ds \paren{C \setminus \set x} \cup \set x$ Set Difference over Subset $\ds$ $=$ $\ds C \cup \set x$ Set Difference with Union $\ds$ $=$ $\ds C$ Union with Superset is Superset

Hence $C = X$ by definition of set equality.

$X \ne C$

Hence:

$x \notin X$
$X \cap \set x = \O$

We have:

 $\ds X$ $=$ $\ds X \setminus \set x$ Set Difference with Disjoint Set $\ds$ $\subseteq$ $\ds C \setminus \set x$ Set Difference over Subset

By definition of set equality:

$X = C \setminus \set x$

It follows that $C \setminus \set x$ is a maximal independent subset of $C$ by definition.

$\blacksquare$