Equivalence of Definitions of Matroid Rank Axioms/Lemma 2
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Theorem
Let $S$ be a finite set.
Let $\rho : \powerset S \to \Z$ be a mapping from the power set of $S$ to the integers.
Let $\rho$ satisfy the rank axioms:
\((\text R 1)\) | $:$ | \(\ds \map \rho \O = 0 \) | |||||||
\((\text R 2)\) | $:$ | \(\ds \forall X \in \powerset S \land y \in S:\) | \(\ds \map \rho X \le \map \rho {X \cup \set y} \le \map \rho X + 1 \) | ||||||
\((\text R 3)\) | $:$ | \(\ds \forall X \in \powerset S \land y, z \in S:\) | \(\ds \map \rho {X \cup \set y} = \map \rho {X \cup \set z} = \map \rho X \implies \map \rho {X \cup \set y \cup \set z} = \map \rho X \) |
Then:
- $\forall A \subseteq S: \map \rho A \le \card A$
Proof
Aiming for a contradiction, suppose:
- $\exists A \subseteq S : \map \rho A > \card A$
Let:
- $A_0 \subseteq S : \card{A_0} = \min \set{\card A : \map \rho A > \card A}$
We have:
\(\ds \map \rho \O\) | \(=\) | \(\ds 0\) | Rank axiom $(\text R 1)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \card \O\) | Cardinality of Empty Set |
Hence:
- $A_0 \ne \O$
Let $y \in A_0$.
From Cardinality of Set Difference:
- $\card {A_0 \setminus \set y} = \card{A_0} - 1 < \card {A_0}$
We have:
\(\ds \map \rho {A_0}\) | \(=\) | \(\ds \map \rho {\paren {A_0 \setminus \set y} \cup \set y}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \map \rho {A_0 \setminus \set y} + 1\) | Rank axiom $(\text R 2)$ | |||||||||||
\(\ds \) | \(\le\) | \(\ds \card {A_0 \setminus \set y} + 1\) | By choice of $A_0$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \card {A_0} - \card{\set y} + 1\) | Cardinality of Set Difference with Subset | |||||||||||
\(\ds \) | \(=\) | \(\ds \card {A_0} - 1 + 1\) | Cardinality of Singleton | |||||||||||
\(\ds \) | \(=\) | \(\ds \card {A_0}\) |
This contradicts the choice of $A_0$:
- $\card{A_0} = \min \set{\card A : \map \rho A > \card A}$
It follows that:
- $\forall A \subseteq S: \map \rho A \le \card A$
$\blacksquare$