Bounds for Rank of Subset

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

Let $\rho : \powerset S \to \Z$ be the rank function of $M$.

Let $A \subseteq S$ be subset of $S$.

Then:
 * $0 \le \map \rho A \le \size A$

Proof
By definition of the rank function:

From Cardinality of Subset of Finite Set:
 * $\forall X \subseteq A : \size X \le \size A$

In particular:
 * $\forall X \subseteq A : X \in \mathscr I$ then $\size X \le \size A$

From Leigh.Samphier/Sandbox/Max yields Supremum of Operands:
 * $\max \set {\size X : X \subseteq A \land X \in \mathscr I} \le \size A$

From Empty Set is Subset of All Sets:
 * $\O \subseteq A$

From Cardinality of Empty Set:
 * $\size \O = 0$

By matroid axiom $(I1)$:
 * $\O \in \mathscr I$

From Leigh.Samphier/Sandbox/Max yields Supremum of Operands:
 * $0 \le \max \set {\size X : X \subseteq A \land X \in \mathscr I}$

It follows that:
 * $0 \le \map \rho A \le \size A$