Leigh.Samphier/Sandbox/Matroid Satisfies Base Axiom/Sufficient Condition/Lemma/Lemma 1

Theoren

Let $U, V$ be finite sets.

Let $\card V < \card U$.

Then:

$\card{V \setminus U} < \card{U \setminus V}$

Proof

We have:

 $\ds \card{U \setminus V}$ $=$ $\ds \card U - \card{U \cap V}$ Cardinality of Set Difference $\ds$ $>$ $\ds \card V - \card{U \cap V}$ As $\card V < \card U$ $\ds$ $=$ $\ds \card{V \setminus U}$ Cardinality of Set Difference

$\blacksquare$