Suppes’ goal: present a system but with a simpler, more intuitive style, suitable for beginners and philosophers. He uses a first-order language with ε (membership) and = (equality), and builds sets from the empty set upward. 2. The Language and Logical Framework Suppes assumes classical first-order logic with identity. The only non-logical primitive is the binary predicate ∈ (membership). All objects are sets—there are no ur-elements (primitive non-set objects). This is a pure set theory .
: The union of two sets is a set.
Denoted ( \bigcup A ). For any set A, there exists a set whose members are exactly all subsets of A. [ \forall A \exists P \forall x [x \in P \leftrightarrow x \subseteq A] ] suppes axiomatic set theory pdf
This avoids Russell’s paradox by restricting comprehension to subsets of existing sets. If a formula ( \phi(x, y) ) defines a functional relation on a set A, then the image of A under that function is a set. This is necessary for constructing ordinals like ( \omega + \omega ) and for proving the existence of ( \aleph_\omega ). Axiom 9: Axiom of Regularity (Foundation) Every non-empty set A has a member disjoint from A. [ \forall A [ A \neq \emptyset \rightarrow \exists x (x \in A \land x \cap A = \emptyset) ] ] Suppes’ goal: present a system but with a
The axioms are intended to be true statements about the cumulative hierarchy of sets, built in stages (ranks). Suppes’ system is essentially Zermelo–Fraenkel without the Axiom of Choice (ZF), though he discusses Choice separately. Below are the core axioms as presented in his book, rephrased for clarity. Axiom 1: Axiom of Extensionality Two sets are equal iff they have the same members. [ \forall x \forall y [ \forall z (z \in x \leftrightarrow z \in y) \rightarrow x = y ] ] The Language and Logical Framework Suppes assumes classical
Suppes’ system is (ZF), plus Choice as an optional axiom. This matches most standard mathematics except for pathological choice-dependent results. 8. Sample Theorem and Proof Style Let’s illustrate Suppes’ rigor with a simple theorem from his book:
Proof : Let ( A ) and ( B ) be sets. By Pairing, ( A, B ) is a set. By Union, ( \bigcup A, B ) is a set. But ( \bigcup A, B = A \cup B ). QED.