# All Bases of Matroid have same Cardinality/Corollary

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:

- $\card X \le \card B$

$\blacksquare$