Element of Matroid Base and Circuit has Substitute

Theorem

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

Let $B \subseteq S$ be a base of $M$.

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

Let $x \in B \cap C$.

Then:

$\exists y \in C \setminus B : \paren{B \setminus \set x} \cup \set y$ is a base of $M$

That is, there exists $y \in C \setminus B$ such that substituting $y$ for $x$ in $B$ is another base.

Proof

By definition of a circuit we have:

Lemma 1

$C \setminus \set x$ is an independent proper subset of $C$

$\Box$

From Independent Subset of Matroid is Augmented by Base:

$\exists X \subseteq B \setminus \paren{C \setminus \set x} : \paren{C \setminus \set x} \cup X$ is a base of $M$

By definition of independent and dependent subsets we have:

Lemma 2

$x \notin \paren {C \setminus \set x} \cup X$

$\Box$

From matroid axiom $(\text I 4)$ we have:

Lemma 3

$\exists y \in \paren {\paren {C \setminus \set x} \cup X} \setminus \paren {B \setminus \set x} : \paren {B \setminus \set x} \cup \set y \in \mathscr I : \card {\paren {B \setminus \set x} \cup \set y} = \card {\paren {C \setminus \set x} \cup X}$

$\Box$

From Independent Subset is Base if Cardinality Equals Cardinality of Base:

$\paren {B \setminus \set x} \cup \set y$ is a base of $M$

Because $x \notin \paren {C \setminus \set x} \cup X$:

$y \ne x$

By definition of set difference:

$y \notin B \setminus \set x$:

By definition of set union:

$y \notin \paren {B \setminus \set x} \cup \set x = B$

By the definition of a subset:

$y \notin X$

By definition of set union:

$y \in C \setminus \set x \subseteq C$

By definition of set difference:

$y \in C \setminus B$

$\blacksquare$