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Inside the constructible universe, every hereditarily countable set appears in a countable level of the constructible hierarchy. Take a countable elementary substructure of a sufficiently large $L_{\eta }$ containing its transitive closure pointwise. The condensation lemma for the constructible universe identifies the collapse with $L_{\beta }$ for countable $\beta$, and the collapse fixes the original set. Conversely, a countable-indexed level is countable inside the constructible universe.

A set is hereditarily countable when its transitive closure is countable. These sets form the transitive set $H_{{\omega }_1}$ and all have rank below ${\omega }_1$.

For an infinite cardinal number $\kappa$, $H_{\kappa }$ consists of the sets $x$ for which $|\mathrm{tc}(\{x\})|<\kappa$, where $\mathrm{tc}$ is transitive closure. The entire transitive membership ancestry is bounded, not only the size of $x$. These collections are transitive sets. At $\kappa ={\omega }_1$ the elements are the hereditarily countable sets.

The structure $H_{\kappa }$ of hereditarily small sets is transitive. If $\kappa \le \omega$, all its sets are hereditarily finite, so it cannot satisfy Infinity. Therefore the assumed model has $\kappa >\omega$.

Let $\mu <\kappa$ be a cardinal. The ordinal $\mu$ belongs to $H_{\kappa }$, and every actual subset of $\mu$ also belongs to $H_{\kappa }$: its transitive closure has size at most $\max (\mu ,\omega )<\kappa$. The Power set axiom inside $H_{\kappa }$ therefore produces the actual $\mathcal{P}(\mu )$, since all subsets relevant to the internal definition are present. This power set itself belongs to $H_{\kappa }$, soThus $\kappa$ is a strong limit. Its regularity was assumed, and we have proved uncountability. Hence $\kappa$ is strongly inaccessible.

The Axiom of foundation states that every nonempty set $A$ has a member $a$ with $a\cap A=\varnothing$. The principle of epsilon induction states that a formula $\varphi$ satisfying $\forall x[(\forall y\in x\varphi (y))\Rightarrow \varphi (x)]$ holds for every set $x$.

Assume Axiom of foundation and suppose a progressive formula has a counterexample $a$. In the transitive closure of $\{a\}$, use axiom schema of separation to collect the counterexamples into a nonempty set $B$. Foundation supplies $b\in B$ having no member in $B$. Since the closure is transitive, every member of $b$ satisfies $\varphi$, contradicting progressiveness. Conversely, if a nonempty set $A$ had no foundation-minimal member, every $x\in A$ would have a member $y\in x\cap A$. The formula $x\notin A$ is then progressive: if every member of $x$ lies outside $A$, $x$ cannot belong to $A$. Epsilon induction would make $A$ empty, a contradiction. This proves equivalence in the presence of the other Zermelo–Fraenkel set theory.

To obtain the cumulative hierarchy, perform epsilon induction on the assertion that $x$ belongs to some $V_{\alpha }$. If this holds for each member $y$, Axiom schema of replacement collects their least such ordinals into a set. Choose an ordinal $\beta$ above them all. Then $x\subseteq V_{\beta }$ and $x\in V_{\beta +1}$. Thus every set occurs in the hierarchy. Use the rank of a set convention $\mathrm{rk}(x)={\sup }_{y\in x}(\mathrm{rk}(y)+1)$, equivalently the least $\alpha$ with $x\subseteq V_{\alpha }$.

The Von Neumann hierarchy and the constructible hierarchy are defined by transfinite recursion:The union clauses apply to nonzero limit ordinals. The definable power set $\mathrm{Def}(L_{\alpha })$ consists of subsets definable over $(L_{\alpha },\in )$ by a first-order formula with finitely many parameters from $L_{\alpha }$. Definability over that structure is essential; it is not unrestricted definability in the ambient universe. The full constructible universe is $L={\bigcup }_{\alpha \in \mathrm{Ord}}{L}_{\alpha }$.

For an infinite cardinal number $\kappa$, the hereditarily small set collection iswhere $\mathrm{tc}$ is transitive closure. Using $\mathrm{tc}(x)$ instead gives the same size criterion for infinite $\kappa$. This bounds the whole membership ancestry, not just $|x|$.

A witness to failure of the Axiom of power set is $\boxed{{\omega }_1\in H_{{\omega }_2}}$. Every subset $A\subseteq {\omega }_1$ belongs to $H_{{\omega }_2}$, since its transitive closure has cardinality at most ${\aleph }_1$. If some $p\in H_{{\omega }_2}$ were the internal power set of ${\omega }_1$, transitivity and bounded subsethood would force $p$ to contain every ambient subset of ${\omega }_1$, and only those subsets. Thus it would be the full $\mathcal{P}({\omega }_1)$.

By the Cantor theorem,Its transitive closure contains all its members, so $|\mathrm{tc}(\{p\})|\ge |p|\ge {\aleph }_2$. This contradicts $p\in H_{{\omega }_2}$. ThereforeThe power-set failure in hereditarily small sets needs no assumption about the Continuum hypothesis.

Use the convention that the transitive closure of $x$ is the least transitive set containing $x$ as a subset. Define $x_0=x$ and $x_{n+1}=\bigcup x_n$. For each natural number $n$, a finite sequence of length $n+1$ with these properties exists by induction, and is unique because each next entry is the uniquely determined union of the previous entry. Thus the relation sending $n$ to its last entry is a definable function-class, not just a putative recursive specification. The Axiom schema of replacement applied to $\omega$ gives the set of all $x_n$. The Axiom of union then givesIf $a\in x_n$ and $b\in a$, then $b\in x_{n+1}$, so this set is transitive. It contains $x=x_0$, and any transitive set containing $x$ contains all $x_n$ by induction; hence it is least. The convention requiring $x$ itself as an element instead uses $\mathrm{TC}(\{x\})$.

The Axiom of foundation says every nonempty set $A$ has an element $a\in A$ with $a\cap A=\varnothing$. Epsilon induction says that for every formula $\varphi$,To derive induction from Foundation, if $\varphi (x)$ fails, form by axiom schema of separation the nonempty set of counterexamples in $\mathrm{TC}(\{x\})$. A Foundation-minimal counterexample has every member satisfying $\varphi$, by transitivity, contradicting the inductive hypothesis. Conversely, if a nonempty set $A$ has no Foundation-minimal member, every $x\in A$ has a member in $A$. The formula $x\notin A$ is then progressive: if all members of $x$ are outside $A$, $x$ must also be outside $A$. Epsilon induction makes $A$ empty, a contradiction. Thus the two principles are equivalent given the other axioms.

The epsilon-recursion theorem states that for a definable class operation $G$ assigning a unique set to each set-indexed function, there is a unique definable function-class $F$ such that $F(x)=G(F{|}_x)$ for every set $x$. A version with explicit $x$ in the argument is equivalent. Foundation gives well-foundedness and Replacement makes each restriction set-sized. Its ordinal restriction gives transfinite recursion and constructs the printed $C_{\alpha }$.

Here “countable” means at most countable, including finite sets and the empty set. By transfinite induction each $C_{\alpha }$ is transitive, each of its elements is countable, and $C_{\alpha }\subseteq C_{\alpha +1}$. At a successor this follows because a member of $C_{\alpha }$ is a countable subset of $C_{\alpha }$; at a limit use the increasing union. Let ${\omega }_1$ be the first uncountable ordinal. Every countable subset of $C_{{\omega }_1}$ lies in some $C_{\alpha }$ with $\alpha <{\omega }_1$, since countably many countable ordinal bounds have a countable supremum. It therefore belongs to $C_{\alpha +1}$. ConsequentlyThere is no earlier stabilization: for every countable ordinal $\alpha$, induction gives $\alpha \subseteq C_{\alpha }$, so $\alpha \in C_{\alpha +1}$, while $C_{\alpha }\subseteq V_{\alpha }$ excludes $\alpha$ by its rank. Thus ${\omega }_1$ is the first such stage. The fixed set is the set of hereditarily countable sets.

Let $\theta =M\cap \mathrm{Ord}$ be the ordinal height of a model of set theory. First $\theta \ge {\omega }_1$. Indeed, if $\theta$ were countable, the internal axiom of choice would give, for every $x\in M$, a bijection in $M$ between $x$ and some ordinal below $\theta$. Such a bijection is also valid externally, making $x$ externally countable. In particular every internal rank $V_{\xi }^M$, $\xi <\theta$, would be countable. Every element of $M$ lies in one of these internal ranks, sowould be a countable union of countable sets, a contradiction. This is why an uncountable transitive set model has uncountable ordinal height.

For $\xi <\theta$, absoluteness of constructible levels gives $L_{\xi }^M=L_{\xi }$, and that level is an element of $M$. The reason is that satisfaction in a fixed set structure uses the same domain and finite formulas in both universes, so successor definitions agree; transfinite recursion then also makes limit stages agree. Transitivity consequently givesIt remains to locate a countability witness. Given an ambient countable ordinal $\alpha$, choose an injection $f:\alpha \to \omega$. Since $V=L$, it belongs to the constructible universe. Its transitive closure, together with $f$ itself, is countable. Choose a countable elementary substructure $X\prec L_{\eta }$ of a sufficiently large limit level containing this closure pointwise. The condensation lemma for the constructible universe identifies the collapse with $L_{\beta }$ for some countable $\beta$. The collapse fixes $f$, since all its hereditary members were included. Thus $f\in L_{\beta }\subseteq L_{{\omega }_1}\subseteq M$.

The ordinal $\alpha$ itself is in $M$, since $\alpha <{\omega }_1\le \theta$. Being an injection between these fixed sets is a bounded formula in set theory, so $M$ recognizes $f$ as a countability witness. This proves countable-ordinal correctness under constructibility:The use of condensation supplies a witness below the height of $M$; merely knowing that $f$ belongs somewhere to $L$ would not be sufficient.