Definition 1.1.1. If \(a\) is a member of \(A,\) we write
\[a\in A.\]This basic relationship is called membership relation. An empty set is a set with no members.
Principle 1.1.2 (Principle of extension). Two objects are equal if they have the same members. Since the same objects have the same members, if follows that, for two objects \(A\) and \(B,\) \(A=B\) if and only if, for all members \(x,\) we have \((x\in A)\Leftrightarrow (x\in B).\)
Theorem 1.1.3. There is only one empty set.
Proof. Let \(E_1\) and \(E_2\) be empty sets. Then there is no element \(x\) that makes the truth values of \(x\in E_1\) and \(x\in E_2\) different from each other. But, according to Principle 1.1.2, two sets can be distinguished only if there exists an element such that \(x\in E_1\) and \(x\notin E_2\) or vice versa. That is, \(E_1=E_2.\)\(\square\)
Notation 1.1.4. We write \(\varnothing\) for the empty set. If \(a_1,\dots,a_n\) are finitely many sets, we write \(\lbrace a_1,\dots,a_n\rbrace\) for \(\lbrace x : x=a_1 \lor \cdots \lor x=a_n\rbrace.\)
Theorem 1.1.5. Everything is a set.
Proof. If \(A\) is an object that is not a set, then there is no \(x\) such that \(x\in A.\) Thus \(A\) is equal to the empty set by Principle 1.1.2, contradicting the assumption that it is not a set. Therefore everything is a set.\(\square\)
Definition 1.1.6. \(A\) is a subset of \(B\) if every element of \(A\) is an element of \(B,\) denoted by \(A\subseteq B.\) That is,
\[A\subseteq B \iff \forall x\ (x\in A \Rightarrow x\in B).\]Then we see that
\[A = B \Leftrightarrow (A\subseteq B \land B\subseteq A).\]If \(A\subseteq B\) and \(A\neq B,\) then \(A\) is a proper subset of \(B,\) denoted by \(A\subset B.\)
Definition 1.1.7. The power set of \(A\) is the set \(\mathcal P(A)\) defined by the equation
\[\mathcal P(A) = \lbrace X: X\subseteq A\rbrace .\]Definition 1.1.8. The union of \(\mathcal A\) is the set \(\bigcup \mathcal A\) defined by the equation
\[\bigcup \mathcal A = \lbrace x : \exists X \in \mathcal A \ (x\in X)\rbrace .\]Then we write \(A\cup B\) for the union of a pair \(\lbrace A, B\rbrace.\) That is,
\[A\cup B = \bigcup \lbrace A, B\rbrace.\]Definition 1.1.9. If \(\mathcal A\) is a nonempty set, the intersection of \(\mathcal A\) is the set \(\bigcap \mathcal A\) defined by the equation
\[\bigcap \mathcal A = \lbrace x : \forall X \in \mathcal A \ (x\in X)\rbrace .\]Then we write \(A\cap B\) for the union of a pair \(\lbrace A, B\rbrace.\) That is,
\[A\cap B = \bigcap \lbrace A, B\rbrace.\]Two sets \(A\) and \(B\) are disjoint if \(A\cap B = \varnothing.\)
References
- Cameron, P. J. (1998). Sets, logic and categories. Springer. https://doi.org/10.1007/978-1-4471-0589-3