Closure of Subset contains Loop

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

Let $x$ be a loop of $M$.

Then:
 * $\forall A \subseteq S : x \in \map \sigma A$

where $\map \sigma A$ denotes the closure of $A$.

Proof
Let $A \subseteq S$.

By definition of the closure of $A$:
 * $x \in \map \sigma A$ $x \sim A$

where $\sim$ is the depends relation on $M$.

By definition of the depends relation:
 * $x \sim A$ $\map \rho {A \cup \set x} = \map \rho A$

where $\rho$ is the rank funtion on $M$.

So it remains to show that:
 * $\map \rho {A \cup \set x} = \map \rho A$

By definition of the rank function:
 * $\map \rho {A \cup \set x} = \max \set{\size X : X \subseteq A \cup \set x \land X \in \mathscr I}$

From Leigh.Samphier/Sandbox/Max Equals an Operand:
 * $\exists X \in \mathscr I : X \subseteq A \cup \set x \land \size X = \map \rho {A \cup \set x}$

From the contrapositive statement of Leigh.Samphier/Sandbox/Set is Dependent if Contains Loop:
 * $x \notin X$

Now:

From Max yields Supremum of Operands:
 * $\size X \le \max \set{\size Y : Y \subseteq A \land Y \in \mathscr I} = \map \rho A$

From Rank Function is Increasing:
 * $\map \rho A \le \map \rho {A \cup \set x} = \size X$

Thus:
 * $\map \rho A = \size X = \map \rho {A \cup \set x}$