## 3.5 The hierarchy of sets

We define by transfinite recursion $V_0 = \emptyset$, $V_{\alpha + 1} = P(V_\alpha )$ (power set), and for a limit ordinal $\alpha$,

$V_\alpha = \bigcup \nolimits _{\beta < \alpha } V_\beta .$

Note that each $V_\alpha$ is a transitive set.

Lemma 3.5.1. Every set is an element of $V_\alpha$ for some ordinal $\alpha$.

Proof. See [Lemma 6.3, Jech]. $\square$

In [Chapter III, Kunen] it is explained that this lemma is equivalent to the axiom of foundation. The rank of a set $S$ is the least ordinal $\alpha$ such that $S \in V_{\alpha + 1}$. By a partial universe we shall mean a suitably large set of the form $V_\alpha$ which will be clear from the context.

## Comments (6)

Comment #6006 by Danny A. J. Gomez-Ramirez on

The definition of rank is not right. The rank of a set S is the least ordinal $\alpha$ such that $S \in V_{\alpha+1}$. Also, note that with the given definition in this section the empty set would have rank 1, which is false. So, for fixing the mistake, you can either change the sub-index by $\alpha+1$, or changing the membership relation $\in$ by the containment relation $\subseteq$ without changing the index.

Comment #6469 by on

The construction is by transfinite recursion not induction. Compare Theorems 2.14 and 2.15 of [Jech]

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