Lemma 3.1.6. (Single choice). Let \(A\) be a non-empty set. Then there exists an objet \(x\) such that \(x\in A\).
\(Proof~by~Tao\). We prove by contradiction. Suppose there does not exist any object \(x\) such that \(x\in A\). Then for all objects \(x\), we have \(x\notin A\). Also, by Axiom 3.2 we have \(x\notin\emptyset\). Thus \(x\in A\iff x\in\emptyset\) (both statements are equally false), and so \(A=\emptyset\) by Definition 3.1.4, a contradiction.