Leigh.Samphier/Sandbox/Matroid with No Circuits Has Single Base

Theorem

Let $M = \struct {S, \mathscr I}$ be a matroid with no circuits.

Then:

$S$ is the only base on $M$.

Proof

From Dependent Subset Contains a Circuit:

$M$ has no dependent subsets

By definition of dependent subsets:

Every subset of $S$ is independent

In particular:

$S \in \mathscr I$

By definition of independent subsets:

$X \in \mathscr I \implies X \subseteq S$

Hence $S$ is a base on $M$ by definition.

Let $X$ be a base on $M$.

Then $X \subseteq S$.

By definition of a base on $M$:

$X$ is a maximal independent subset of $M$

As $S \in \mathscr I$, by definition of a maximal set:

$X = S$

Hence:

$S$ is the only base on $M$.

$\blacksquare$

Sources

• 1976: Dominic Welsh: Matroid Theory ... (previous) ... (next) Chapter $1.$ $\S 9.$ Circuits