Loop Belongs to Every Flat

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

Let $A \subseteq S$.

Let $x \in S$.

If $x$ is a loop and $A$ is a flat subset then $x \in A$.

Proof
We proceed by Proof by Contraposition.

That is, it is shown that:
 * if $x \notin A$ then either $x$ is not a loop or $A$ is a flat subset

Let $x \notin A$.

Let $x$ be a loop.

By definition of a loop:
 * $\set x$ is a dependent subset.

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

Let $X \in \mathscr I$ such that $X \subseteq A \cup \set x$.

From Superset of Dependent Set is Dependent:
 * $\set x \not \subseteq X$

From Singleton of Element is Subset:
 * $x \notin X$

So:
 * $X \subseteq A$

By definition of the rank function:
 * $\size X \le \map \rho A$

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

Then:
 * $\map \rho A = \map \rho {A \cup \set x}$

If follows that $A$ is not a flat subset by definition.

It has been shown that:
 * if $x \notin A$ then either $x$ is not a loop or $A$ is a flat subset

The theorem holds by the Rule of Transposition.