Suppes Axiomatic Set Theory | Pdf

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.

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 ] ] suppes axiomatic set theory pdf

This ensures that a set is determined solely by its elements. There exists a set with no members. [ \exists x \forall y (y \notin x) ]

From this we get singletons (when a = b) and unordered pairs. For any set A, there exists a set whose members are exactly the members of members of A. [ \forall A \exists U \forall x [x \in U \leftrightarrow \exists y (x \in y \land y \in A)] ] Suppes’ system is (ZF), plus Choice as an optional axiom

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) ] ]

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] ] Sample Theorem and Proof Style Let’s illustrate Suppes’

Denoted ( \emptyset ). For any sets a, b, there exists a set whose members are exactly a and b. [ \forall a \forall b \exists x \forall y (y \in x \leftrightarrow y = a \lor y = b) ]

: The union of two sets is a set.