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

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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

From Set Difference is Subset:

$C \setminus \set x \subseteq C$

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

$C \setminus \set x \ne C$

From Leigh.Samphier/Sandbox/Proper Subset of Matroid Circuit is Independent and matroid axiom $(\text I 1)$:

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


Let $X$ be an independent subset such that:

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

As $C$ is dependent:

$X \ne C$


Aiming for a contradiction, suppose:

$x \in X$

From Singleton of Element is Subset:

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

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.

This contradicts the fact that:

$X \ne C$

Hence:

$x \notin X$


From Intersection With Singleton is Disjoint if Not Element:

$X \cap \set x = \O$

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.

$\blacksquare$