My question is: "Is it possible to have a sound and rigorous legitimation of the following construction ?". This construction is:

0/ Let ZFC be the usuel set theory, and let us add to the language capital latin letters as names for classes.Let V={x/x=x} be the usual universal class (that is a proper class, directly by foundation, or by Russel's paradox and separation) and P(x)={y/y⊆x} be the usual Power set fonction.

1/ Let us now extend the domain of the power set fonction to all classes, so that P(A)={y/y⊆A} We have that P(A)is a set if A is a set by the power set axiom;but as F(x)={x} in an injection from A into P(A), we also hav}e that P(A) must be a proper class if A is such. So that P(A) is a set iff A is a set. The universal class V being transitive and well-founded, we have that V={x/x⊆V}={x/x⊆V}=P(V); So that: V=P(0,V)=P(V)=P(1,V)=P(P(n,V))=P(n+1,V). Let us also remark that for every class A, we have A⊆V, so tat V is the greater possible class.

2/ Now, let us recursively build the following Q construction: Q(0,V)=P(V)=V{x/x⊆V}=V and Q(n+1,V)={A/(B∈A-->B∈Q(n,V)}={A/A Q(n,V)} Moreover let the collection Q be the union of the collections Q(n,V): Q=∪〈n∈N〉Q(n,V);

By induction, we have Q(n,V)⊆Q(n+1,V). We already know that the elements of Q(0,V)=V are axactly sets, and that the elements of Q(1,V)={A/A⊆V} are exactly usual classes. So that Q(0,V)⊆Q(1,V); moreover, the elements of the difference Q(1,V)/Q(0,V) are exactly usual proper classes and V is such an element. We also have Q(n,V)⊆Q(n+1,V)-->Q(n+1,V)⊆Q(n+2,V) because A∈Q(n+1,V)<-->(∀B∈A(B∈Q(n,V)) and by hypothesis B∈Q(n,V)-->B∈Q(n+1,V), so that A∈Q(n+2,V).

It is now natural to name sets as level 0 super-classes, proper classes as level 1 super-classes, and more generally elements of the difference Q(n+1,V)/Q(n,V) as level (n+1) super-classes, and also members of Q as super-classes.

We already know that Q(0,V) and Q(1,V)/Q(0,V) are non-void. more generally we have that Q(n+1,V)/Q(n,V) is non-void. To see this, let A and B be two level n super-classes and consider R(A,B)={D∈Q(n,V)/A⊆D⊆B}. It is clear that R(A,B) is a level (n+1) super-class that is non-void as soon as A⊆B. Particularly R(∅,A)={B/B⊆A} and R(A,A)={A} are such level (n+1) super-classes. So, we have by induction that Q(n,V)⊆Q(n+1,V) and also that Q(n,V)∈Q(n+1,V); for every level n, Q(n,V) is the greater element of Q(n+1,V) for inclusion.

3/ We also obtain that the difference of two super-classes (of level n) is a super-class (of level n), and that the union (the intersection) of a finite (of any family, but what can be the collection of indices ?) family of super-classes (of level n) is a super-class (of level n; of level ≤n).

Gérard Lang

  • 3
    $\begingroup$ Please edit this question to add paragraphs and MathJax. It is almost unreadable in its current form. $\endgroup$ – Michael Greinecker Sep 30 '12 at 15:03
  • 2
    $\begingroup$ What Michael said, sans "almost". $\endgroup$ – Asaf Karagila Sep 30 '12 at 15:16
  • $\begingroup$ I agree that the current form of the text is (almost or without almost) unreadable. This was not the case of my original text, that had many paragraphs and was structured (notably by 0/, 1/,2/ and 3/) so as to be as readable as possible ! Gérard Lang $\endgroup$ – Gérard Lang Sep 30 '12 at 15:55
  • 2
    $\begingroup$ You may want to get yourself better acquainted with the way the site parses the text. For example a single linebreak is ignored, much like in LaTeX, so if you wish to start a new paragraph you should use a double linebreak. Similarly, there is a nice support of basic LaTeX commands via $ symbols (and $$ for centered display math), but you need to sometimes add spaces, or use \lbrace instead of \{, to avoid conflicts with the parser. $\endgroup$ – Asaf Karagila Sep 30 '12 at 16:08

One of the easiest ways to justify your construction is this:

Take an inaccessible cardinal $\kappa$. Then $V_\kappa$, the $\kappa$'th iterate of the powerset operation starting from the empty set and taking unions at limit steps, is a model of ZFC. Now consider a two-sorted structure with the underlying sets $V_\kappa$ and $V_{\kappa+\omega}\setminus V_\kappa$. The elements of $V_\kappa$ are your "sets", the elements of $V_{\kappa+\omega}\setminus V_\kappa$ are the superclasses. If you want to be able to talk about the different levels, add unary predicates for the individual levels, or even a binary predicate that is satisfied by a superclass $C$ and $n$ if $C$ is a superclass of the $n$-th level.

Does this help?

  • $\begingroup$ Yes,thank you,this helps me very much, even if this wis not was I was dreaming of ! In fact, I hoped that there was some possible comparison with Ackermann's set theory with foundation as developped by Azriel Lévy, and further by Muller. $\endgroup$ – Gérard Lang Sep 30 '12 at 16:06
  • $\begingroup$ After reflexion, it is clear that your model satisfies all the specifications of my construction. So maybe it fulfills most of my dream; anyway, I credit your answer, even if I would like some more comments. Gérard Lang $\endgroup$ – Gérard Lang Oct 2 '12 at 15:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.