Uniform Matroid is Matroid

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Theorem

Let $S$ be a finite set of cardinality $n$.

Let $0 \le k \le n$.

Let $U_{k, n} = \struct{S, \mathscr I}$ be the uniform matroid of rank $k$.


Then $U_{k, n}$ is a matroid.

Proof

It needs to be shown that $\mathscr I$ satisfies the matroid axioms $(I1)$, $(I2)$ and $(I3)$.

Matroid Axiom $(I1)$

Now:

\(\ds \size \O\) \(=\) \(\ds 0\) Cardinality of Empty Set
\(\ds \) \(\le\) \(\ds k\) Assumption


By definition of the uniform matroid of rank $k$:

$\O \in \mathscr I$

It follows that $\mathscr I$ satisfies matroid axiom $(I1)$.

$\Box$

Matroid Axiom $(I2)$

Let $X \in \mathscr I$.

Let $Y \subseteq X$.

Then:

\(\ds \size Y\) \(\le\) \(\ds \size X\) Cardinality of Subset of Finite Set
\(\ds \) \(\le\) \(\ds k\) Definition of uniform matroid of rank $k$


By definition of the uniform matroid of rank $k$:

$Y \in \mathscr I$

It follows that $\mathscr I$ satisfies matroid axiom $(I2)$.

$\Box$

Matroid Axiom $(I3)$

Let $U, V \in \mathscr I$.

Let $\size V < \size U$.


From Intersection is Subset:

$U \cap V \subseteq V$
$U \cap V \subseteq U$

Then:

\(\ds \size {U \cap V}\) \(\le\) \(\ds \size V\) Cardinality of Subset of Finite Set
\(\ds \) \(<\) \(\ds \size U\) Assumption

Then:

$U \cap V \ne U$

It follows that:

$U \cap V \subsetneqq U$


Then:

\(\ds \exists x\) \(\in\) \(\ds U \setminus {U \cap V}\) Set Difference with Proper Subset
\(\ds \) \(=\) \(\ds U \setminus V\) Set Difference with Intersection is Difference


Then:

\(\ds \size {V \cup \set x}\) \(=\) \(\ds \size V + \size {\set x}\) Corollary to Cardinality of Set Union
\(\ds \) \(=\) \(\ds \size V + 1\) Cardinality of Singleton
\(\ds \) \(\le\) \(\ds \size U\) Assumption
\(\ds \) \(\le\) \(\ds k\) Definition of uniform matroid of rank $k$


By definition of the uniform matroid of rank $k$:

$V \cup \set x \in \mathscr I$

It follows that $\mathscr I$ satisfies matroid axiom $(I3)$.

$\blacksquare$

Sources