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
\(C\) 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,
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 \(\mathcal A.\)
Theorem 3.1.9. Let \(A,B,\) and \(C\) be sets. Then following holds:
- \(A\cap (B \cup C) = (A\cap B) \cup (A \cap C).\)
- \(A\cup (B \cap C) = (A\cup B) \cap (A \cup C).\)
Proof.
(1) Suppose that \(x\in A\cap (B \cup C).\) Then \(x\in A\) and \(x\in (B\cup C).\) This is divided into two cases. \(x\in A\) and \(x\in B,\) the first case, and \(x\in A\) and \(x\in B,\) the second case. In the first case, \(x\in (A\cap B)\) so that \(x\in (A\cap B) \cup (A \cap C).\) In the second case, \(x\in (A\cap C)\) so that \(x\in (A\cap B) \cup (A \cap C).\) Thus, if \(x\in A\cap (B \cup C),\) then \(x\in (A\cap B) \cup (A \cap C).\)
Conversely, we suppose that \(x\in (A\cap B) \cup (A \cap C).\) In the same way, we can see that if \(x\in (A\cap B) \cup (A \cap C)\) then \(x\in A\cap (B \cup C).\)
Therefore if follows that \(A\cap (B \cup C)\) and \((A\cap B) \cup (A \cap C)\) are bound to consist of the same elements.
(2) It is proven in the same way as above. \(\square\)
Theorem 3.1.10. Let \(A,B,\) and \(C\) be sets. Then following holds:
- \(A- (B \cup C) = (A- B) \cup (A - C).\)
- \(A- (B \cap C) = (A- B) \cap (A - C).\)
Proof. It is proven in the same way as in the Theorem 3.1.9.
Axiom 3.1.11 (Power set).
\[\forall X\, \exists \mathcal P\,\forall Y\ (Y\in \mathcal P \leftrightarrow Y\subseteq X). \tag{\(\text{ZF5}\)}\]Definition 3.1.12. 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 3.1.13. 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
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