Equivalence of Definitions of Matroid Circuit Axioms/Lemma 4
![]() | 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:
\((\text C 1)\) | $:$ | \(\ds \O \notin \mathscr C \) | |||||||
\((\text C 2)\) | $:$ | \(\ds \forall C_1, C_2 \in \mathscr C:\) | \(\ds C_1 \ne C_2 \implies C_1 \nsubseteq C_2 \) | ||||||
\((\text C 3')\) | $:$ | \(\ds \forall C_1, C_2 \in \mathscr C:\) | \(\ds C_1 \ne C_2 \land z \in C_1 \cap C_2 \land w \in C_1 \setminus C_2 \implies \exists C_3 \in \mathscr C : w \in C_3 \subseteq \paren {C_1 \cup C_2} \setminus \set z \) |
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:
- $\map t {x_1, \ldots, x_q} = \ds \sum_{i = 1}^q \map \theta {x_1, \ldots, x_q}_i$
Let $\rho : \powerset S \to \Z$ be the mapping satisfying the matroid rank axioms defined by:
- $\forall A \subseteq S$:
- $\map \rho A = \begin{cases} 0 & : \text{if } A = \O \\ \map t {x_1, \ldots, x_q } & : \text{if } A = \set{x_1, \ldots, x_q} \end{cases}$
Let $M = \struct{S, \mathscr I}$ be the matroid corresponding to the rank function $\rho$.
Let $\mathscr C_M$ be the set of circuits of $M$
Then:
- $\forall C \in \mathscr C_M : \exists C' \in \mathscr C : C' \subseteq C$
Proof
Let $C \in \mathscr C_M$.
By definition of circuit:
- $C$ is dependent
By matroid axiom $(\text I 1)$:
- $C \neq \O$
Let $x \in C$.
From Proper Subset of Matroid Circuit is Independent:
- $C \setminus \set x$ is independent
We have:
\(\ds \map \rho C\) | \(=\) | \(\ds \card C - 1\) | Rank of Matroid Circuit is One Less Than Cardinality | |||||||||||
\(\ds \) | \(=\) | \(\ds \card C - \card{\set x}\) | Cardinality of Singleton | |||||||||||
\(\ds \) | \(=\) | \(\ds \card {C \setminus \set x}\) | Cardinality of Set Difference | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \rho {C \setminus \set x}\) | Rank of Independent Subset Equals Cardinality |
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 $\exists C \in \mathscr C : y \in C \subseteq X \cup \set y$
$\Box$
From Lemma 2:
- $\exists C' \in \mathscr C : x \in C' \subseteq C$
$\blacksquare$