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 Leigh.Samphier/Sandbox/Matroid with Dependent Subsets Has Circuits:
 * $M$ 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 $M$.