User:Leigh.Samphier/Matroids/Dual of Dual Matroid Equals Matroid

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Theorem

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

Let $M^*$ denote the dual of $M$.


Then:

$\paren{M^*}^* = M$


That is, the dual $\paren{M^*}^*$ of $M^*$ is $M$.


Proof

Let $\mathscr B$ denote the set of bases of $M$.

Let $\paren{\mathscr B^*}^*$ denote the set of bases of $\paren{M^*}^*$.


By definition of dual matroid:

We have:

\(\ds \paren{\mathscr B^*}^*\) \(=\) \(\ds \set{S \setminus \paren{S \setminus B} : B \in \mathscr B}\) Definition of Dual Matroid
\(\ds \) \(=\) \(\ds \set{B : B \in \mathscr B}\) Complement of Complement
\(\ds \) \(=\) \(\ds \mathscr B\)


Let $\paren{\mathscr I^*}^*$ denote the independent subsets of $\paren{M^*}^*$.


We have:

\(\ds \paren{\mathscr I^*}^*\) \(=\) \(\ds \set{X \subseteq S : \exists B \in \mathscr B : X \subseteq B}\) User:Leigh.Samphier/Matroids/Characterization of Matroid Independent Sets in Terms of Bases
\(\ds \) \(=\) \(\ds \mathscr I\) User:Leigh.Samphier/Matroids/Characterization of Matroid Independent Sets in Terms of Bases


By definition of matroid:

$\paren{M^*}^* = M$

$\blacksquare$


Sources

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