Element of Matroid Base and Circuit has Substitute/Lemma 2

Lemma for Element of Matroid Base and Circuit has Substitute
Let $M = \struct {S, \mathscr I}$ be a matroid.

Let $C \subseteq S$ be a dependent subset of $M$.

Let $x \in C$.

Let $X \subseteq S$ such that:
 * $\paren {C \setminus \set x} \cup X \in \mathscr I$

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

Proof

 * $x \in \paren {C \setminus \set x} \cup X$
 * $x \in \paren {C \setminus \set x} \cup X$

Then:
 * $\set x, C \setminus \set x \subseteq \paren {C \setminus \set x} \cup X$

From Union of Subsets is Subset:
 * $C \subseteq \paren {C \setminus \set x} \cup X$

From matroid axiom $(\text I 2)$:
 * $C \in \mathscr I$

This contradicts:
 * $C \notin \mathscr I$

It follows that:
 * $x \notin \paren {C \setminus \set x} \cup X$