3.1. Axioms of Set Theory
Definition 3.1.1. For convenience, we first introduce a syntactic sugar: If \(\varphi (x,p_1, \dots, p_n)\) is a formula, we call
\[C = \lbrace x : \varphi (x,p_1, \dots, p_n)\rbrace\]a class. Members of \(C\) are all those sets \(x\) that satisfy \(\varphi (x,p_1, \dots, p_n):\)
\[x\in C \leftrightarrow \varphi (x,p_1, \dots, p_n).\]Every set can be represented as a class. If \(S\) is a set, the class \(\lbrace x : x\in S\rbrace\) denotes the set \(S\) itself. If \(C = \lbrace x : \varphi (x,p_1, \dots, p_n) \rbrace\) and \(D = \lbrace x : \psi (x,q_1, \dots, q_m)\rbrace\) then
\[\forall x \ (\varphi (x,p_1, \dots, p_n) \leftrightarrow \psi (x,q_1, \dots, q_m)) \rightarrow C = D.\]\(C\) is a subclass of \(D\) if and only if for all \(x,\) \(x\in C\) implies \(x\in D,\) denoted by \(C\subseteq D.\)
Definition 3.1.2. We define the following operations on classes:
\[\begin{align} C \cap D &=\lbrace x : x\in C \land x \in D\rbrace,\\ C \cup D &=\lbrace x : x\in C \lor x\in D\rbrace,\\ C - D &=\lbrace x : x\in C \land x\notin D\rbrace,\\ \bigcup C &=\lbrace x : \exists X \in C \ (x\in X)\rbrace. \end{align}\]Axiom 3.1.3 (Extensionality).
\[\forall X\, \forall Y \ (\forall x\ (x \in X \leftrightarrow x \in Y) \rightarrow X = Y). \tag{\(\text{ZF1}\)}\]Axiom 3.1.4 (Pairing).
\[\forall x\, \forall y\, \exists X\, \forall z \ (z \in X \leftrightarrow (z = x \vee z = y)). \tag{\(\text{ZF2}\)}\]By Extensionality, the set \(X\) is unique.
Definition 3.1.5. For two sets \(a\) and \(b,\) the pair \(\lbrace a,b\rbrace\) is defined by
\[\lbrace a,b\rbrace = \lbrace x : x=a \lor x=b \rbrace.\]We define an ordered pair to be the set \((a,b)\) such that
\[(a,b) = \lbrace \lbrace a \rbrace , \lbrace a,b\rbrace \rbrace .\]We further define ordered triples, quadruples, etc., as follows:
\[\begin{aligned} (a,b,c) &= ((a,b),c), \\ (a,b,c,d) &= ((a,b,c),d), \\ &\ \ \vdots \\ (a_1,\dots,a_{n+1}) &= ((a_1,\dots,a_n),a_{n+1}). \end{aligned}\]Axiom 3.1.6 (Separation schema). Let \(\varphi (x,p)\) be a formula.
\[\forall X\, \forall p \, \exists Y \,\forall x\ (x\in Y\leftrightarrow x\in X\land \varphi (x,p)). \tag{\(\text{ZF3}\)}\]This can be generalized:
\[\forall X\, \forall p_1 \dots \forall p_n \,\exists Y \,\forall x\ (x\in Y\leftrightarrow x\in X\land \varphi (x,p_1,\dots, p_n)).\]We can express this by using a class: Let \(C = \lbrace x : \varphi (x,p_1, \dots, p_n) \rbrace,\) then
\[\forall X\, \exists Y \ (X\cap C = Y).\]It follows that the empty class \(\varnothing = \lbrace x : x\neq x\rbrace\) is a set—the empty set.
Definition 3.1.7. By the separation axioms, the intersection and the difference of two sets are sets. So we define the set operations
\[X \cap Y = \lbrace x\in X : x \in Y\rbrace \quad \text{and} \quad X-Y = \lbrace x\in X : x\notin Y\rbrace.\]If \(A\cap B=\varnothing,\) two sets \(A\) and \(B\) are disjoint. If \(C\) is a nonempty class,
\[\bigcap C = \lbrace x : \forall X\ (X \in C \rightarrow x\in X)\rbrace\]is a set.
Axiom 3.1.8 (Union).
\[\forall \mathcal X\, \exists X\, \forall x \ (x \in X \leftrightarrow \exists Y\ (Y \in \mathcal X \land x \in Y )). \tag{\(\text{ZF4}\)}\]Thus, for a set \(\mathcal A\) there is a unique set
\[A = \lbrace x : \exists B\ (B\in \mathcal A \land x\in B)\rbrace = \bigcup \mathcal A,\]the union of \(\mathcal A.\)
Axiom 3.1.9 (Power set).
\[\forall X\, \exists \mathcal P\,\forall Y\ (Y\in \mathcal P \leftrightarrow Y\subseteq X). \tag{\(\text{ZF5}\)}\]Definition 3.1.10. Let \(A\) be a set. The power set of \(A\) is the set \(\mathcal P(A)\) defined by
\[\mathcal P(A) = \lbrace X : X\subseteq A\rbrace.\]Definition 3.1.11. We use the abbreviation \(\lbrace (x,y) : \varphi(x,y)\rbrace\) for \(\lbrace z : \exists x\, \exists y\ (z=(x,y)\land \varphi(x,y))\rbrace.\) Let \(A\) and \(B\) be sets. We define the product of \(A\) and \(B\) to be the class such that
\[A\times B = \lbrace (x,y) : x\in A \land x\in B\rbrace.\]If \((x,y)\in A\times B\) then there exist \(x\in A\) and \(y\in B\) so that
\[\lbrace x\rbrace , \lbrace x,y\rbrace \in \mathcal P(A\cup B).\]Thus
\[(x,y)=\lbrace \lbrace x\rbrace , \lbrace x,y\rbrace \rbrace \subseteq \mathcal P(A\cup B).\]Therefore \((x,y)\in \mathcal P (\mathcal P(A\cup B)).\) It follows that \(A\times B\subseteq \mathcal P (\mathcal P(A\cup B)).\) Consequently the product \(A\times B\) is a set. We further define \(A\times B\times C = (A\times B)\times C,\) and in general
\[A_1\times \cdots \times A_{n+1} = (A_1\times \cdots \times A_{n}) \times A_{n+1}.\]We also let
\[A^n = \underbrace{A \times \cdots \times A}_{n\ \text{times}}.\]References
- Jech, T. (2003). Set theory (3rd ed.). Springer-Verlag. https://doi.org/10.1007/3-540-44761-X