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
- 1976: Dominic Welsh: Matroid Theory ... (previous) ... (next) Chapter $1.$ $\S 6.$ Properties of the rank function