Transfinite recursion and the rank hierarchy

Suppose a fixed formula defines a class operation F assigning a unique set to every set-length function that can appear as a history. Transfinite recursion gives, for each ordinal θ, a unique function h with domain θ whose value at α is F applied to h restricted to α. Proving uniqueness uses the first stage where two proposed histories disagree. Existence is a different argument: compatible partial histories are assembled, and Replacement ensures that the required collection at each set-sized limit is a set.

The rule may be presented through separate zero, successor, and limit clauses, provided those clauses specify unique sets and agree with the claimed history dependence. The theorem does not allow an output to be an arbitrary proper class. Nor does it assert that the function on all ordinals is one set. Instead the compatible set-length solutions define a class function, described by a formula. In ZF this is a scheme for definable rules; arbitrary class parameters require a separately stated class theory.

Set V at stage zero equal to the empty set. At a successor stage take the power set of the preceding stage; at a nonzero limit take the union of all earlier stages. The limit collection is legitimate because the index ordinal is a set and the earlier-stage operation is definable and functional. Each stage is transitive, and earlier stages are included in later ones. These facts follow by transfinite induction, with Power Set and Union doing different work in the successor and limit arguments.

Foundation permits the rank assignment: the rank of a set is the supremum of the successors of the ranks of its members. A well-founded recursion on the membership relation within the transitive closure of the set establishes the assignment without asking for a set of all sets. Every set belongs to a stage of the hierarchy. More precisely x belongs to Vα exactly when its rank is below α; x is a subset of Vα exactly when its rank is at most α. The empty case is consistent with both statements.

A singleton can have high rank if its sole member already has high rank. Conversely, an infinite set such as ω has rank ω rather than a finite number determined by the number of membership steps in one chosen element. Rank is therefore different from cardinality and different from the time used by an algorithm representing the set. When calculating it, apply the successor before taking the supremum; omitting that successor shifts the hierarchy membership statement by one stage.

Finite stages make the distinction inspectable. The first stage has one element, the second has two, and the third has four; the next has sixteen. These are iterated power-set sizes, not successive additions of one point. At the limit ω, the union contains exactly the hereditarily finite sets. It is infinite even though each earlier stage is finite. A finite picture can verify initial ranks and recursive clauses, but it cannot establish the existence of the full hierarchy or justify replacing a class recursion by a set.