Finite level super classes over ZFC - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T18:54:39Z http://mathoverflow.net/feeds/question/108460 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108460/finite-level-super-classes-over-zfc Finite level super classes over ZFC Gérard Lang 2012-09-30T11:49:27Z 2013-02-09T13:55:17Z <p>My question is: "Is it possible to have a sound and rigorous legitimation of the following construction ?". This construction is:</p> <p>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.</p> <p>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.</p> <p>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);</p> <p>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)&lt;-->(∀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).</p> <p>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.</p> <p>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.</p> <p>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).</p> <p>Gérard Lang</p> http://mathoverflow.net/questions/108460/finite-level-super-classes-over-zfc/108471#108471 Answer by Stefan Geschke for Finite level super classes over ZFC Stefan Geschke 2012-09-30T14:15:26Z 2012-09-30T14:15:26Z <p>One of the easiest ways to justify your construction is this:</p> <p>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.</p> <p>Does this help?</p>