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Use the standard convention that a strongly inaccessible cardinal is uncountable, regular and strong limit. The printed explanatory definition omits uncountability: literally it also admits $\kappa =\omega$, for which no finite cardinal is a worldly cardinal. Thus that omission makes the requested conclusion false under the literal abbreviated definition. The proof below applies to the usual, intended strongly inaccessible cardinal convention $\kappa >\omega$.

First $|V_{\gamma }|<\kappa$ for every $\gamma <\kappa$. At successor stages this follows from the strong limit cardinal property, and at limits from the regular cardinal property. Consequently $V_{\kappa }\models \mathsf{ZFC}$. In particular, for Axiom schema of replacement, a domain $a\in V_{\kappa }$ has size less than $\kappa$, so its functional image has fewer than $\kappa$ elements, so the supremum of the rank of a set over its elements is below $\kappa$ by regular cardinal structure. All other axioms, including axiom of choice, have their witnesses at bounded ranks below $\kappa$; uncountability supplies axiom of infinity.

Enumerate the first-order formulas as ${\phi }_0,{\phi }_1,\dots$. For each finite collection, reflection inside the set structure $V_{\kappa }$ gives a closed unbounded subset of $\kappa$ of agreeing ranks. To see why the witness bounds stay below $\kappa$, there are fewer than $\kappa$ parameter tuples at each $V_{\gamma }$, and the supremum of their least witness ranks remains below $\kappa$ by regular cardinal structure. Closing under these bounds and taking increasing countable limits gives the usual reflection theorem for definable hierarchies argument. The countable intersection of these closed unbounded subsets is still closed unbounded because $\kappa$ is regular and uncountable. At each resulting $\alpha$,The infinite cardinal numbers below $\kappa$ also form a closed unbounded subset: they are unbounded because ${\lambda }^{+}\le 2^{\lambda }<\kappa$ for infinite $\lambda <\kappa$, and a supremum of increasing cardinal numbers is a cardinal number. Intersect the two closed unbounded subsets. Every member of the intersection is a worldly cardinal, and an unbounded subset of a regular cardinal $\kappa$ has size $\kappa$. HenceIn fact this proves the stronger worldly cardinals below an inaccessible cardinal result that these worldly cardinals contain a closed unbounded subset of $\kappa$.

A definable continuous hierarchy is a definable class function $\alpha \mapsto H_{\alpha }$ on the ordinals, with set-valued levels, such that $H_{\alpha }\subseteq H_{\beta }$ for $\alpha <\beta$ andfor nonzero limit ordinals. Its union is a definable class in set theory $H$. Often the definition additionally requires every level to be a transitive set; the following statement also works without that requirement.

The reflection theorem for definable hierarchies says that for any finite collection $\mathcal{F}$ of first-order formulas, there is a closed unbounded class of ordinals $\alpha$ such that, for every $\phi \in \mathcal{F}$ and every parameter tuple from $H_{\alpha }$,A useful justification is the witness-closure proof. Close $\mathcal{F}$ under subformulas. At each level and for each existential subformula, bound the least level containing a witness for each parameter tuple for which a witness exists in $H$. Axiom schema of replacement bounds these indices. Iterating the finitely many bounds through $\omega$ produces a limit level containing all required witnesses. Induction on the first-order formulas, equivalently the finite-formula version of the Tarski-Vaught test, yields agreement. Continuity gives closedness of the reflecting class, and starting above any prescribed ordinal gives unboundedness. This is finite reflection, not a claim that all first-order formulas reflect simultaneously in an arbitrary hierarchy.

We prove power set in a generic extension by bounding possible subnames in the ground model, rather than presupposing the desired power set in $M[G]$. Let $x={\tau }^G$, and let $S=\{\rho :\exists r((\rho ,r)\in \tau )\}$. In $M$ formEvery $\nu \in B$ is a forcing name. Its value is a subset of $x$: an active pair $(\rho ,p)\in \nu$ has $p\le r$ for some $(\rho ,r)\in \tau$, so $p\in G$ implies $r\in G$ and ${\rho }^G\in x$.

Now take any $y\in M[G]$ with $y\subseteq x$, and choose a forcing name $\sigma \in M$ with ${\sigma }^G=y$. By axiom schema of separation and definability of the syntactic forcing relation,The atomic membership truth lemma for forcing shows ${\nu }_{\sigma }^G=y$. One inclusion follows immediately from its soundness direction. For the other, if $a\in y\subseteq x$, choose $(\rho ,r)\in \tau$ with $r\in G$ and ${\rho }^G=a$. The truth direction supplies $p\in G$ forcing $\rho \in \sigma$; strengthen within $G$ below $p$ and $r$ to obtain an active pair in ${\nu }_{\sigma }$ representing $a$.

Finally the ground-model forcing nameexists by Axiom schema of replacement and the ground-model power set axiom. Since $G$ is nonempty, all $\nu \in B$ contribute their values, andThis set belongs to $M[G]$ by the definition of a generic extension. The construction uses all conditions in the outer pairs and therefore does not require a greatest condition in $\mathbb{P}$.

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.

Use standard notation for forcing, so $q\le p$ means that $q$ is stronger. Let $\check{p}$ be the canonical forcing name for the ground-model condition $p$. DefineHere $q\perp p$ means that they are incompatible forcing conditions. The checks and their collection are formed by recursion and Axiom schema of replacement in $M$, and the displayed set is selected by axiom schema of separation, so this is a forcing name in $M$.

By evaluation of a forcing name,If $p\in G$, directedness of the generic filter gives a common stronger condition for $p$ and each $q\in G$, so $p$ is absent from this value.

Conversely, for fixed $p\in \mathbb{P}$ the setis a dense subset of a forcing order belonging to $M$. A condition incompatible with $p$ is already in it, and one compatible with $p$ has a common extension below $p$. Genericity supplies $q\in G\cap D_p$. If $p\notin G$, upward closure of $G$ rules out $q\le p$, hence $q\perp p$. ThereforeThis forcing name for the complement of a generic filter works without a separativity assumption on the order.

Fix $a\in L$. Inside the ambient transitive model $M$, the Axiom of power set makes the collection of constructible subsets of $a$ a set. For each such subset $x$, choose the least stage of the constructible hierarchy at which $x$ appears. The Axiom schema of replacement and the supremum of a set of ordinals give an ordinal $\beta$ bounding all these stages; enlarge $\beta$ so that $a\in L_{\beta }$.

Nowis definable over $L_{\beta }$ with parameter $a$. It therefore belongs to the definable power set $\mathcal{D}(L_{\beta })=L_{\beta +1}\subseteq L$. This set contains exactly the subsets of $a$ that belong to the constructible universe, so it witnesses the Axiom of power set in $L$. Therefore $L\models$ Power Set.

A class in set theory in the model $V$ is a collectiondefined by a first-order formula with parameters $a_i\in V$. A set-theoretic class function is a definable class relation $F$ for which every input in its domain has exactly one output. Informally, the Axiom schema of replacement says that the image $F‘‘a$ of any set $a$ under any such function class is again a set.

Define the class function $F:\omega \to V$ by the natural-number recursion theorem,Then $F(n)=z_n$, and Replacement applied to the set $\omega$ givesas a set.

Call a set small when it injects into some $z_n$. Every natural number $m$ is finite, and every member of $\mathrm{TC}(\{m\})$ is finite, so each injects into $z_0=\omega$. Hence $m\in \mathbf{HS}$. The set $\omega$ itself injects into $z_0$, and its hereditary members are natural numbers, so $\omega \in \mathbf{HS}$.

We next prove $z_n\in \mathbf{HS}$ by mathematical induction. The case $z_0=\omega$ was just proved. If $z_n\in \mathbf{HS}$ and $u\in z_{n+1}=\mathcal{P}(z_n)$, then $u\subseteq z_n$, so inclusion injects $u$ into $z_n$ and makes $u$ small. Every set below $u$ in its transitive closure is already below $z_n$ and is small by the induction hypothesis. The set $z_{n+1}$ itself injects into $z_{n+1}$. Thus every member of $\mathrm{TC}(\{z_{n+1}\})$ is small.

By Cantor theorem, $|z_{n+1}|>|z_n|$, so the $z_n$ are distinct and $z_n\mapsto n$ injects $y$ into $\omega =z_0$. Thus $y$ is small. Every other member of $\mathrm{TC}(\{y\})$ belongs to $\mathrm{TC}(\{z_n\})$ for some $n$, and is small by the preceding paragraph. Therefore $y\in \mathbf{HS}$. This proves the finite-power-set hereditary-small construction.

The structure $(\mathbf{HS},\in )$ is not a model of ZF because it fails the Axiom of union. If $\bigcup y$ were small, it would inject into some $z_m$. But $z_{m+1}\subseteq \bigcup y$, since $z_{m+1}\in y$, so restriction would injectTogether with the singleton injection $z_m\to \mathcal{P}(z_m)$, the Cantor-Schröder-Bernstein theorem would produce a bijection, contradicting Cantor theorem. Hence $\bigcup y$ is not small and therefore does not belong to $\mathbf{HS}$. Since $y\in \mathbf{HS}$ has no union inside the class, the Union axiom fails.

Fix a set $x$ and define by recursion on $\omega$The recursion theorem defines a function-class $F$ uniquely by $F(n)=T_n$: for each $n\in \omega$ there is exactly one finite sequence satisfying the displayed recursion through stage $n$. Thus $F$ is genuinely functional, and the Axiom schema of replacement applied to the set $\omega$ gives the set $\{T_n:n\in \omega \}$.

Now defineIt contains every member of $x$. If $y\in T$, then $y\in T_n$ for some $n$; hence every $z\in y$ belongs toThus $T$ is transitive. If $U$ is any transitive set containing every member of $x$, induction gives $T_n\subseteq U$ for every $n$, so $T\subseteq U$. Thereforeis the transitive closure of $x$.

Work in the ambient universe and let $a,u\in L$. SupposeFor each $x\in a$, let $y_x$ be this unique witness. The relativization ${\phi }^L$ is a first-order formula, so the ambient Axiom schema of replacement collects the witnesses $y_x$ into a set. Every witness lies in the constructible hierarchy, hence there is an ordinal $\alpha$ such thatFor example, take the supremum of one constructible rank for each witness and then increase it by one.

The set $L_{\alpha }$ itself belongs to $L_{\alpha +1}\subseteq L$. Taking $b=L_{\alpha }$, every $x\in a$ has a witness $y\in b$ satisfying ${\phi }^L(x,y,u)$. Thereforewhich is the stated instance of Replacement.

Let $a,p\in M[G]$, and suppose a first-order formula $\psi (x,y,p)$ defines exactly one $y$ for every $x\in a$. Apply the Lévy reflection theorem to the formulas needed to express this assertion, choosing an ordinal $\alpha$ with $a,p\in V_{\alpha }^{M[G]}$ such thatfor every $x\in a$. Therefore every required value lies in the set $V_{\alpha }^{M[G]}$.

The already established axiom schema of separation forms the setFunctionality makes $b$ exactly the range of the definable function on $a$. This proves every instance of the Axiom schema of replacement in the generic extension $M[G]$.

A finite collection of first-order formulas has a closed unbounded class of ordinals whose levels in a definable continuous hierarchy agree with its union for all parameters in the level. Close the collection under subformulas, bound the least witness levels for existential instances on each set-sized stage by Axiom schema of replacement, and iterate these bounds countably. Continuity supplies the closed stages, and induction on formulas proves agreement. The Tarski-Vaught test describes the same witness criterion. For a hierarchy of length an uncountable regular cardinal, bounds stay below that cardinal when every stage has smaller cardinality.

To separate a subset of a constructible set by a formula interpreted in the constructible universe, apply the reflection theorem for definable hierarchies to the formula and its subformulas. At a sufficiently large reflecting level containing the parameters, the desired subset is definable and belongs to the next constructible hierarchy level. The reflection proof uses least witness stages and ambient Axiom schema of replacement, so does not presuppose Axiom schema of separation within the constructible universe.

Let $M$ be a transitive set model of ZFC. If $\mathrm{Ord}\cap M$ were countable, then every $x\in M$ would be countable: the internal axiom of choice supplies a bijection from $x$ to an ordinal of $M$, which is externally countable. For every $\alpha \in \mathrm{Ord}\cap M$, the internal rank $V_{\alpha }^M$ is an element of $M$ and hence countable. The Axiom schema of replacement inside $M$ gives $M={\bigcup }_{\alpha \in \mathrm{Ord}\cap M}{V}_{\alpha }^M$, a countable union of countable sets. Thus every uncountable transitive set model has uncountably many ordinals.