# Independent Set can be Augmented by Larger Independent Set/Corollary

< Independent Set can be Augmented by Larger Independent Set(Redirected from Independent Subset of Matroid is Augmented by Base)

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## Theorem

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

Let $X \subseteq S$ be an independent subset of $M$.

Let $B \subseteq S$ be a base of $M$.

Then:

- $\exists Z \subseteq B \setminus X : \card{X \cup Z} = \card B : X \cup Z$ is a base of $M$

## Proof

From Cardinality of Independent Set of Matroid is Smaller or Equal to Base:

- $\card X \le \card B$

### Case 1: $\card X < \card B$

From Independent Set can be Augmented by Larger Independent Set:

- $\exists Z \subseteq B \setminus X : X \cup Z \in \mathscr I : \card {X \cup Z} = \card B$

$\Box$

### Case 2: $\card X = \card B$

Let $Z = \O$.

Then:

- $Z \subseteq B \setminus X : X \cup Z \in \mathscr I : \card {X \cup Z} = \card B$

$\Box$

From Independent Subset is Base if Cardinality Equals Cardinality of Base:

- $X \cup Z$ is a base of $M$

$\blacksquare$