Leigh.Samphier/Sandbox/Rank of Matroid Circuit is One Less Than Cardinality

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Theorem

Let $M = \struct {S, \mathscr I}$ be a matroid.

Let $C \subseteq S$ be a circuit of $M$.

Let $\rho: \powerset S \to \Z$ denote the rank function of $M$.


Then:

$\map \rho C = \card C -1$

Proof

By definition of a circuit:

$C$ is dependent

By matroid axiom $(\text I 1)$:

$C \ne \O$


Let $x \in C$.


Lemma

$C \setminus \set x$ is a maximal independent subset of $C$


We have:

\(\ds \map \rho C\) \(=\) \(\ds \card{C \setminus \set x}\) Leigh.Samphier/Sandbox/Cardinality of Maximal Independent Subset Equals Rank of Set
\(\ds \) \(=\) \(\ds \card C - \card{\set x}\) Cardinality of Set Difference with Subset
\(\ds \) \(=\) \(\ds \card C - 1\) Cardinality of Singleton

$\blacksquare$

Sources