User:Leigh.Samphier/Matroids/Equivalence of Definitions of Matroid Rank Axioms
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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.
The following definitions for the Rank Axioms are equivalent:
Formulation 1
$\rho$ is said to satisfy the rank axioms if and only if
\((\text R 1)\) | $:$ | \(\ds \map \rho \O = 0 \) | |||||||
\((\text R 2)\) | $:$ | \(\ds \forall X \in \powerset S \land y \in S:\) | \(\ds \map \rho X \le \map \rho {X \cup \set y} \le \map \rho X + 1 \) | ||||||
\((\text R 3)\) | $:$ | \(\ds \forall X \in \powerset S \land y, z \in S:\) | \(\ds \map \rho {X \cup \set y} = \map \rho {X \cup \set z} = \map \rho X \implies \map \rho {X \cup \set y \cup \set z} = \map \rho X \) |
Formulation 2
$\rho$ is said to satisfy the rank axioms if and only if
\((\text R 4)\) | $:$ | \(\ds \forall X \in \powerset S:\) | \(\ds 0 \le \map \rho X \le \size X \) | ||||||
\((\text R 5)\) | $:$ | \(\ds \forall X, Y \in \powerset S:\) | \(\ds X \subseteq Y \implies \map \rho X \le \map \rho Y \) | ||||||
\((\text R 6)\) | $:$ | \(\ds \forall X, Y \in \powerset S:\) | \(\ds \map \rho {X \cup Y} + \map \rho {X \cap Y} \le \map \rho X + \map \rho Y \) |
Proof
Formulation 1 implies Formulation 2
Follows immediately from:
$\Box$
Formulation 2 implies Formulation 1
Let $\rho$ satisfy formulation 2 of the rank axioms:
\((\text R 4)\) | $:$ | \(\ds \forall X \in \powerset S:\) | \(\ds 0 \le \map \rho X \le \size X \) | ||||||
\((\text R 5)\) | $:$ | \(\ds \forall X, Y \in \powerset S:\) | \(\ds X \subseteq Y \implies \map \rho X \le \map \rho Y \) | ||||||
\((\text R 6)\) | $:$ | \(\ds \forall X, Y \in \powerset S:\) | \(\ds \map \rho {X \cup Y} + \map \rho {X \cap Y} \le \map \rho X + \map \rho Y \) |
$\rho$ satisfies $(\text R 1)$
We have:
\(\ds 0\) | \(\le\) | \(\ds \map \rho \O\) | Rank axiom $(\text R 4)$ | |||||||||||
\(\ds \) | \(\le\) | \(\ds \card \O\) | Rank axiom $(\text R 4)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 0\) | Cardinality of Empty Set |
Hence:
- $\map \rho \O = 0$
$\Box$
$\rho$ satisfies $(\text R 2)$
Let $X \subseteq S$.
Let $y \in S$.
We have:
\(\ds \map \rho X\) | \(\le\) | \(\ds \map \rho {X \cup y}\) | Rank axiom $(\text R 5)$ | |||||||||||
\(\ds \) | \(\le\) | \(\ds \map \rho X + \map \rho {\set y} - \map \rho {X \cup \set y}\) | Rank axiom $(\text R 6)$ | |||||||||||
\(\ds \) | \(\le\) | \(\ds \map \rho X + \map \rho {\set y}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \map \rho X + \card {\set y}\) | Rank axiom $(\text R 4)$ | |||||||||||
\(\ds \) | \(\le\) | \(\ds \map \rho X + 1\) | Cardinality of Singleton |
This proves rank axiom $(\text R 2)$
$\Box$
$\rho$ satisfies $(\text R 3)$
Let $X \subseteq S$.
Let $x, y \in S$.
Let $\map \rho {X \cup \set x} = \map \rho {X \cup \set y} = \map \rho X$.
We have:
\(\ds \map \rho X\) | \(\le\) | \(\ds \map \rho {X \cup \set x \cup \set y }\) | Rank axiom $(\text R 5)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \rho {\paren{X \cup \set x} \cup \paren{X \cup \set y} }\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \map \rho {X \cup \set x} + \map \rho {X \cup \set x} - \map \rho {\paren{X \cup \set x} \cap \paren{X \cup \set y} }\) | Rank axiom $(\text R 6)$ | |||||||||||
\(\ds \) | \(\le\) | \(\ds \map \rho {X \cup \set x} + \map \rho {X \cup \set x} - \map \rho X\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \rho X + \map \rho X - \map \rho X\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \rho X\) |
Hence:
- $\map \rho {X \cup \set x \cup \set y} = \map \rho X$
This proves rank axiom $(\text R 3)$
$\blacksquare$
Sources
- 1976: Dominic Welsh: Matroid Theory Chapter $1.$ $\S 6.$ Properties of the rank function
- 2018: Bernhard H. Korte and Jens Vygen: Combinatorial Optimization: Theory and Algorithms (6th ed.) Chapter $13$ Matroids $\S 13.2$ Other Matroid Axion, Theorem $13.10$