All Bases of Matroid have same Cardinality/Corollary
< All Bases of Matroid have same Cardinality(Redirected from Cardinality of Independent Set of Matroid is Smaller or Equal to Base)
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 $M = \struct {S, \mathscr I}$ be a matroid.
Let $B \subseteq S$ be a base of $M$.
Let $X \subseteq S$ be any independent subset of $M$.
Then:
- $\card X \le \card B$
Proof
From Independent Subset is Contained in Maximal Independent Subset :
- $\exists B' \subseteq S : X \subseteq B'$ and $B'$ is a maximal independent subset of $S$
By definition of a base:
- $B'$ is a base of $M$
From Cardinality of Subset of Finite Set:
- $\card X \le \card {B'}$
From All Bases of Matroid have same Cardinality:
- $\card{B'} = \card B$
Hence:
- $\card X \le \card B$
$\blacksquare$