All Bases of Matroid have same Cardinality/Corollary

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:

$\card X \le \card B$

$\blacksquare$