Axiom 3.2. (Empty set). There exists a set \(\emptyset\), known as the empty set, which contains no elements, i.e., for every object \(x\) we have \(x\notin\emptyset\).
Axiom 3.2
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Lemma 3.1.6
\(Proof~by~Tao\). 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.
- Exercise 3.2.1
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Exercise 3.1.2
Exercise 3.1.2 Using only Definition 3.1.4, Axiom 3.1, Axiom 3.2, and Axiom 3.3, prove that the sets \(\emptyset, \{\emptyset\}, \{\{\emptyset\}\}\), and \(\{\emptyset, \{\emptyset\}\}\) are all distinct (i.e, no two of them are equal to each other).
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Chapter3
Exercise 3.1.2 Using only Definition 3.1.4, Axiom 3.1, Axiom 3.2, and Axiom 3.3, prove that the sets \(\emptyset, \{\emptyset\}, \{\{\emptyset\}\}\), and \(\{\emptyset, \{\emptyset\}\}\) are all distinct (i.e, no two of them are equal to each other).