# Independent Subset is Base if Cardinality Equals Rank of Matroid/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 $M$.

Let $\card X = \card B$.

Then:

- $X$ is a base of $M$.

## Proof

From All Bases of Matroid have same Cardinality:

- $\card B = \map \rho S$

where $\rho$ denotes the rank function on $M$.

Hence:

- $\card X = \map \rho S$

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

- $X$ is a base of $M$.

$\blacksquare$