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

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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

$\blacksquare$