Bounds for Rank of Subset

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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:

\(\ds \map \rho A\) \(=\) \(\ds \max \set {\size X : X \subseteq A \land X \in \mathscr I}\)


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 Max Operation 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 $(\text I 1)$:

$\O \in \mathscr I$

From Max Operation Yields Supremum of Operands:

$0 \le \max \set {\size X : X \subseteq A \land X \in \mathscr I}$


It follows that:

$0 \le \map \rho A \le \size A$

$\blacksquare$


Sources