Equivalence of Definitions of Matroid Circuit Axioms/Condition 2 Implies Condition 4

From ProofWiki
Jump to navigation Jump to search


This article needs proofreading.
Please check it for mathematical errors.
If you believe there are none, please remove {{Proofread}} from the code.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Proofread}} from the code.


Theorem

Let $S$ be a finite set.

Let $\mathscr C$ be a non-empty set of subsets of $S$ that satisfies the circuit axioms:

\((\text C 1)\)   $:$   \(\ds \O \notin \mathscr C \)      
\((\text C 2)\)   $:$     \(\ds \forall C_1, C_2 \in \mathscr C:\) \(\ds C_1 \ne C_2 \implies C_1 \nsubseteq C_2 \)      
\((\text C 3')\)   $:$     \(\ds \forall C_1, C_2 \in \mathscr C:\) \(\ds C_1 \ne C_2 \land z \in C_1 \cap C_2 \land w \in C_1 \setminus C_2 \implies \exists C_3 \in \mathscr C : w \in C_3 \subseteq \paren {C_1 \cup C_2} \setminus \set z \)      


Then:

$\mathscr C$ is the set of circuits of a matroid $M = \struct{S, \mathscr I}$ on $S$


Proof

We will define a mapping $\rho$ associated with $\mathscr C$.

It will be shown that $\rho$ is the rank function of a matroid $M$ which has $\mathscr C$ as the set of circuits.


For any ordered tuple $\tuple{x_1, \ldots, x_q}$ of elements of $S$, let $\map \theta {x_1, \ldots, x_q}$ be the ordered tuple defined by:

$\forall i \in \set{1, \ldots, q} : \map \theta {x_1, \ldots, x_q}_i = \begin{cases} 0 & : \exists C \in \mathscr C : x_i \in C \subseteq \set{x_1, \ldots, x_i}\\ 1 & : \text {otherwise} \end{cases}$


Let $t$ be the mapping from the set of ordered tuple of $S$ defined by:

$\map t {x_1, \ldots, x_q} = \ds \sum_{i = 1}^q \map \theta {x_1, \ldots, x_q}_i$

Lemma 1

Let $\tuple{x_1, \ldots, x_q}$ be any ordered tuple of elements of $S$.

Let $\pi$ be any permutation of $\tuple{x_1, \ldots, x_q}$.


Then:

$\map t {x_1, \ldots, x_q} = \map t {x_{\map \pi 1}, \ldots, x_{\map \pi q}}$

$\Box$


Let $\rho : \powerset S \to \Z$ be the mapping defined by:

$\forall A \subseteq S$:
$\map \rho A = \begin{cases} 0 & : \text{if } A = \O \\ \map t {x_1, \ldots, x_q } & : \text{if } A = \set{x_1, \ldots, x_q} \end{cases}$


From Lemma 1:

$\rho$ is well-defined


We now show that $\rho$ satisfies the rank function axioms:

\((\text R 1)\)   $:$   \(\ds \map \rho \O = 0 \)      
\((\text R 2)\)   $:$     \(\ds \forall X \in \powerset S \land y \in S:\) \(\ds \map \rho X \le \map \rho {X \cup \set y} \le \map \rho X + 1 \)      
\((\text R 3)\)   $:$     \(\ds \forall X \in \powerset S \land y, z \in S:\) \(\ds \map \rho {X \cup \set y} = \map \rho {X \cup \set z} = \map \rho X \implies \map \rho {X \cup \set y \cup \set z} = \map \rho X \)      


$\rho$ satisfies $(\text R 1)$

Follows from the definition of $\rho$.

$\Box$

$\rho$ satisfies $(\text R 2)$

Let:

$X = \set{x_1, \ldots, x_q} \subseteq S$ and $y \in S$


We have:

\(\ds \map \rho {X \cup \set y} - \map \rho X\) \(=\) \(\ds \map t {\tuple{x_1, \dots, x_q, y} } - \map t {\tuple{x_1, \dots, x_q} }\) Definition of $\rho$
\(\ds \) \(=\) \(\ds \map \theta {\tuple{x_1, \dots, x_q, y} }_{q+1}\) Definition of $t$
\(\ds \) \(=\) \(\ds \begin{cases} 0 & : \exists C \in \mathscr C : y \in C \subseteq \set{x_1, \ldots, x_q, y}\\ 1 & : \text {otherwise} \end{cases}\) Definition of $\theta$
\(\ds \leadstoandfrom \ \ \) \(\ds 0\) \(\le\) \(\ds \map \rho {X \cup \set y} - \map \rho X \le 1\)
\(\ds \leadstoandfrom \ \ \) \(\ds \map \rho X\) \(\le\) \(\ds \map \rho {X \cup \set y} \le \map \rho X + 1\) Adding $\map \rho X$ to all terms

$\Box$


Lemma 2

Let $X \subseteq S$ and $y \in S \setminus X$.


Then:

$\map \rho {X \cup \set y} = \map \rho X$ if and only if $\exists C \in \mathscr C : y \in C \subseteq X \cup \set y$

$\Box$


$\rho$ satisfies $(\text R 3)$

Let:

$X \subseteq S$ and $y, z \in S$

Let:

$\map \rho {X \cup \set y} = \map \rho {X \cup \set z} = \map \rho X$
Case 1 : $z \in X$

Let $z \in X$.

Then:

$\map \rho {X \cup \set {y,z}} = \map \rho {X \cup \set y}$

Since $\map \rho {X \cup \set y} = \map \rho X$, then:

$\map \rho {X \cup \set {y,z}} = \map \rho X$

$\Box$

Case 2 : $z = y$

Let $z = y$.

Then

$\map \rho {X \cup \set {y,z}} = \map \rho {X \cup \set y}$

Since $\map \rho {X \cup \set y} = \map \rho X$, then:

$\map \rho {X \cup \set {y,z}} = \map \rho X$

$\Box$

Case 3 : $z \neq y$ and $z \notin X$

Let $z \neq y$ and $z \notin X$.

Then $z \notin X \cup \set y$.

From Lemma 2:

$\exists C_z \in \mathscr C : z \in C_z \subseteq X \cup \set{z} \subseteq X \cup \set{y,z}$

From Lemma 2:

$\map \rho {X \cup \set{y, z}} = \map \rho {X \cup \set y} = \map \rho X$

$\Box$


From Equivalence of Definitions of Matroid Rank Axioms:

$\rho$ is the rank function of a matroid $M = \struct{S, \mathscr I}$ on $S$

Lemma 3

$\mathscr C$ is the set of circuits of $M$.

$\blacksquare$

Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Equivalence_of_Definitions_of_Matroid_Circuit_Axioms/Condition_2_Implies_Condition_4&oldid=613074"