Let a be a set. Set Comprehension can be applied to x ⊆ a, because any x satisfying this is a subclass ofaand hence a set.
Separation
Letabe a set and ϕbe an∈-formula. x∈a∧ϕM(x) defines a class, because allx satisfying this are ina and hence sets. This class is a subclass of aand hence a set.
Union
∃z (x ∈ z∧z ∈ a) defines a set, because any x satisfying it is a set by two applications of Element.
Infinity
Applying Class Comprehension toL(x)∧C(x), one can establish that there is a classV containing all sets and nothing else. Now we can apply Set Compre-hension to ϕ(x) := ∀I (∅ ∈ I∧ ∀y (y ∈ I → y∪ {y} ∈ I)→ x∈I), because
∅ ∈V ∧ ∀y(y∈V →y∪ {y} ∈V), i.e. anyxsatisfyingϕ(x) is inV and hence a set.
Remark. The sets that we have shown to exist do not only satisfy their exten-sionality conditions for sets, but for any objects. For example, the doubleton {a, b} established by pairing satisfies not only ∀x(set(x)→(x∈ {a, b} ↔x= a∨x=b)) but also the stronger statement ∀x(x∈ {a, b} ↔x=a∨x=b).
This strong characterization of these sets is usually needed when we apply these axiom in what follows.
We use the following standard definitions:
Definition 4.1.13. Theordered pair (x, y) is defined to be{{x},{x, y}}if this class exists.
Remark. Given Pairing and the Extensionality Axiom ofAU, (x, y) always exists for limitedx, y. If xor y is unlimited, it is possible that (x, y) does not exist (though it might turn out to exist even in that case). An atomic statement involving (x, y) should be considered false if (x, y) does not exist. The same convention holds for all other terms we define without proving that they exist in all cases.
Definition 4.1.14. Arelation is a class of ordered pairs. Given a relation R, we writeR(x, y) for (x, y)∈R.
Definition 4.1.15. A map is a relation R such that R(x, y1) and R(x, y2) implies y1 = y2. Given a map f, we write f(x) for the element y such that (x, y)∈f if such ayexists.
Definition 4.1.16. For a relationR, thedomain ofR(dom(R)) is the class of xsuch that∃y R(x, y) if such a class exists.
Definition 4.1.17. A class xis transitive (trans(x)) iff for allz ∈ y ∈ x, we havez∈x.
Definition 4.1.18. A classxiswell-ordered by∈iff∀y, z(y /∈y∧(y∈z∨z∈ y∨y=z)∧ ∀u⊆x(u6=∅ → ∃t(t∈u∧@s(s∈t∧s∈u)))).
As noted by L´evy and Vaught (1961) (pages 1054-1055), the expected defini-tion of ordinals as transitive classes well-ordered by∈does not suffice to prove that any two ordinals are comparable. Hence they added the condition that for any two subclasses x, yof an ordinal, x\y should exist. Since we are now working in a theory with urelements, we have to add the additional condition that all elements of an ordinal are classes:
Definition 4.1.19. x is an ordinal (Ord(x)) iff x is a transitive class well-ordered by∈,∀y, z ⊆x∃u∀t(t∈u↔t∈y∧t /∈z) and∀y∈x C(y).
Definition 4.1.20. xis anordinal number (ord(x)) iff xis an ordinal and is limited.
Definition 4.1.21. For an ordinal x,x0 denotesx∪ {x} if this class exists.
Definition 4.1.22. Whenxandyare ordinals,x < yis an alternative notation forx∈y.
Lemma 4.1.23. An element of an ordinal is an ordinal.
Proof. Letxbe an ordinal andy∈x. The transitivity and∈-well-orderedness of ycan be established using standard techniques. yis a class by the last condition in the definition ofOrd(x). Every element ofy is an element ofxand hence a class. Finally, leta andb be subclasses of y. These are subclasses of x, so by Ord(x)a\bexists as required.
Lemma 4.1.24. For any AU-formulaϕ(x), the following is a theorem of AU: If ∃x(ϕ(x)∧ord(x)), then there is a leastxsuch that ϕ(x)∧ord(x).
Proof. Chooseysuch thatϕ(y)∧ord(y). Ify is minimal with this property, we are done, so assume it is not. Sinceyis a set,{x∈y |ϕ(x)}defines a set. Since it is a non-empty subset ofy andyis well-ordered, it has a minimal elementx.
If there is az < xsuch thatϕ(z), thenz∈y by transitivity ofy, contradicting the choice ofx. Soxis minimal such thatϕ(x)∧ord(x).
Remark. This allows us to give proofs by transfinite induction over the ordinal numbers (but not over all ordinals).
Lemma 4.1.25. If α and β are ordinals, then precisely one of the following properties holds:
α < β
β < α
α=β
Proof. The fact that at most one of these properties holds is easily proved.
Now suppose for a contradiction that none of these three properties holds.
α\β andβ\αexist by the additional condition imposed on ordinals.
Suppose for a contradiction that α\β is empty, i.e. α⊆β. Since α6=β, β\αis non-empty. Letxbe the minimal element ofβ\α.
4.1. ACKERMANN SET THEORY 55 Supposey ∈x. Then by the transitivity ofβ, y∈β. Ify were not inα, it would be inβ\α, contradicting the minimality of x. Soy∈α. Thusx⊆α.
Conversely, suppose y ∈α. Theny ∈β. Furthermorey 6=x, since x /∈ α.
Additionallyx /∈y, for otherwise we would havex∈αby the transitivity ofα.
Since xand y are both in β, which is totally ordered by ∈, we may conclude thaty∈x. Thusα⊆x.
So x=α. But since x∈β, we can now conclude that α∈β, contrary to our assumption.
Thus we have thatα\β is non-empty. Similarly,β\αis non-empty. Let a be the minimal element ofα\β and letbbe the minimal element of β\α.
Letx∈a. By minimality ofa,x∈β. Ifx=borb∈x, then the transitivity of αimplies thatb ∈ α, contrary to the choice of b. Since ∈ totally orders β, x∈b. Thusa⊆b.
Similarly,b⊆a, i.e.a=b. But thenb∈α, contradicting the choice ofb.
Lemma 4.1.26. The union of a set of ordinal numbers is an ordinal number.
Proof. The existence of this union follows from the Axiom of Union that we have proved. Its transitivity and ∈-well-orderedness can be established using standard techniques. It clearly contains only classes. And since subclasses of it are sets, their subtraction certainly exists by Separation.
Since we have not proved Replacement, we cannot use transfinite recursion.
Nevertheless, the von-Neumann hierarchy of Vα’s can be shown to exist for ordinal numbersα(but not for any ordinal α, though for some it does exist).
We introduce a notation for speaking about theVα’s and about restrictions of the mapα7→Vα without ontological commitment:
Definition 4.1.27. Given an ordinal α, V•|α denotes the map such that dom(V•|α) = α and ∀x ∈ α (y ∈ Vx|α ↔ ∃z ∈ x y ⊆ Vz|α), if such a map exists and is unique. (HereVx|α is a convenient notation forV•|α(x).)
Definition 4.1.28. Given an ordinalα,Vαdenotes the classVα|α0 if this class exists.
Lemma 4.1.29. For every ordinal numberα,Vαexists and is a set.
Proof. If for all ordinal numbers β, V•|β0 were a set, then the lemma would clearly hold. So assume for a contradiction thatβis the smallest ordinal number such thatV•|β0 is not a set. Then the formula∃x < β(z=V•|x0) holds only for setsz, so by Set Comprehensionu:={z| ∃x < β z=V•|x0}is a set. SoV•|β= Suis a set. But thenVβ ={x| there is a pair (y, z)∈V•|β such thatx⊆z}
is a set by Set Comprehension. Hence V•|β0 =V•|β∪(β, Vβ) is a set, contrary to our assumption.
Definition 4.1.30. xis apure set (pset(x)) iff∃α(ord(α)∧x∈Vα).
Definition 4.1.31. xis a pure class (PC(x)) iff C(x)∧ ∀y ∈x∃α(Ord(α)∧ trans(Vα)∧ ∀z∈VαC(z)∧y∈Vα).
Remark. Even though we can show by transfinite induction that for all ordinal numbersα,Vα is transitive and contains only classes, this cannot be shown for ordinals, since we cannot carry out transfinite induction over the ordinals. So the conditionstrans(Vα) and∀z∈VαC(z) do make a difference.
Definition 4.1.32. For a pure set x, the rank of x (rank(x)) is the smallest ordinal numberαsuch thatx∈Vα.
Remark. By Lemma 4.1.24,rank(x) is well-defined for all pure setsx.
Definition 4.1.33. For anA∗-formulaϕ, we define the translationϕpto be the AU-formula obtained by replacing all occurrences of M bypsetand restricting all quantifiers byPC.
Definition 4.1.34. For a set Φ of A∗-formulae, define Φp:={ϕp |ϕ∈Φ}.
Now the following theorem establishes thatAU interpretsA∗: Theorem 4.1.35. AU`A∗p.
Proof. We have to prove that for every axiomϕofA∗,ϕpcan be proved inAU: Extensionality
We have to prove that∀x, y(PC(x)∧PC(y)∧ ∀z(PC(z)→(z∈x↔z∈y))→ x=y). Since all pure classes are classes, AU-Extensionality implies that it is enough to show that any element of a pure class is a pure class. Letxbe a pure class and let y ∈ x. Then there is an ordinal α such thatVα is transitive, all elements of Vα are classes andy ∈ Vα. Then y is a class, and if z ∈ y, then z∈Vαby transitivity ofVα, soy is a pure class as required.
Class Comprehension
Given anyA∗-formulaϕsuch that for allx,ϕp(x) implies thatxis a pure set, we have to prove that there is a pure classysuch that a pure classzis inyiffϕp(z).
We apply AU’s Class Comprehension to ϕp(x). The resulting class is a pure class, because all of its elements are pure sets, i.e. in aVαfor an ordinal number α(and as remarked above, for ordinal numbersα, the constraintstrans(Vα) and
∀z∈VαC(z) certainly hold).
Set Comprehension
Given any ∈-formulaϕsuch that for all x, ϕp(x) implies that xis a pure set, we have to prove that there is a pure set y such that a pure class z is in y iff ϕb(z). Since ϕp(x) does not contain the symbolL, we may applyAU’s Set Comprehension toϕp(x) to show thaty:={x|ϕp(x)} is a set.
Now “z is the rank of an element of y” holds only for sets. So v :=
{z | zis the rank of an element of y} is a set. But then µ := S
v is an ordi-nal number by Lemma 4.1.26. Nowy ⊆Vµ, i.e.y∈Vµ0, i.e.y is a pure set, as required.
Element Axiom
An element of a pure set is a pure set, since ∀α (ord(α) → trans(Vα) ∧
∀z∈VαC(z)) by transfinite induction.
Subset Axiom
We need to show that a subclass of a pure set is a pure set. This directly follows from the fact that for an ordinal numberα,Vα is closed under subclasses.