Replacement axiom and the von Neumann hierarchy Within ZFC, the von Neumann hierarchy consists of sets $V_\alpha$ indexed by ordinals, subject to the following conditions:

*

*$V_0=\varnothing$.

*$V_{\alpha+1}=\mathcal P(V_\alpha)$.

*$V_\lambda=\bigcup_{\beta<\lambda}V_\beta$ for limit $\lambda$.

My question is: what is the formal justification for the last step?
The axiom of union would allow us to construct $V_\lambda$ if we already had a set $\{V_\beta:\beta<\lambda \}$. However, the existence of $V_\beta$ for $\beta<\lambda$ does not obviously guarantee the existence of this set: $V_{\omega\cdot 2}$ in ZF famously models itself minus the axiom of replacement. This of course suggests that with replacement, this set could be constructed as the image of $\lambda$ (which under the von Neumann ordinal construction is the set of all lower ordinals) under the class function $V$.
But then $V$ would have to be a class function somehow defined through transfinite induction. Since predicates can't refer to themselves, and since we can't just assert the existence and uniqueness of a class function as a theorem the way we can with normal functions, the way this works eludes me.
 A: This is really just a long comment, but the phrasing on the wikipedia page for replacement is verbose and perhaps obscuring how to use it here. Consider this version of replacement:

For any set $X$ and binary predicate $\phi(-,-)$ such that for each element $x\in X$ there exists a unique set $y_x$ such that $\phi(x,y_x)$ is true, there exists a set $Y$ whose members are precisely the sets $y_x$ such that there exists some $x\in X$ with $\phi(x,y_x)$ true. We denote the set $Y$ guaranteed by this axiom together with a set $X$ and binary predicate $\phi(-,-)$ by $$\{y_x:x\in X\}.$$

This is equivalent to all other standard phrasings of replacement over the rest of the $ZFC$ axioms, and is easier to use in the situation you outiline above. Specifically, take $X=\lambda$ and let $$\phi(-,-)=\text{All sets up to rank $-$ are members of $-$, and nothing else.}$$ which accepts ordinals in the first argument and arbitrary sets in the second. For each $\beta<\lambda$ we have that $\phi(\beta,V_\beta)$ is true, and if $\phi(\beta,Z)$ is true for some other set $Z$ then $Z=V_\beta$ by extensionality, so $V_\beta$ is unique satisfying $\phi(\beta,V_\beta)$ for all $\beta<\lambda$. Consequently $$\{V_\beta:\beta\in\lambda\}$$ is a set by replacement, and the union of this set is $V_\lambda$.
A: The other answer does not actually give you $V$ all at once. But it is in fact easy!
$
\def\empty{\varnothing}
\def\pow{\mathcal{P}}
$
Let $FN(S) = \{ \ f : ∀x{∈}S\ ∃!y ( \ ⟨x,y⟩∈f \ ) \ \}$, namely the class of all functions on $S$.
Let $SUCC = \{ \ succ(k) : k{∈}ORD \ \}$, namely the class of all successor ordinals.
Let $R(f) = \cases{ 
f\cup\{⟨k,\bigcup_{i{∈}k} f(i)⟩\} & if $f{∈}FN(k)$ for some $k{∈}ORD{∖}SUCC$ 
\\ f\cup\{⟨succ(k),\pow(f(k))⟩\} & if $f{∈}FN(succ(k))$ for some $k{∈}ORD$ }$ (and note that we do not need to treat the zero ordinal case separately).
Let $V = \{ \ ⟨k,R(f)⟩ : k{∈}ORD ∧ f{∈}FN(k) ∧ ∀i{∈}k\ ( \ f(i)=R(f{↾}i) \ ) \ \}$, namely the class of all pairs that each corresponds to the 'limit' of a function that satisfies the recursive relation $R$.
Then you can prove that $V$ satisfies the desired properties. In particular, if there is any $k{∈}ORD$ such that $¬∃!x\ ( \ ⟨k,x⟩∈V \ )$, then there is an $∈$-mininum such $k$, and you can easily obtain a contradiction.
The point is that we can define this $V$ even without replacement, and only need replacement to prove that it is a class function on $ORD$. Also, the technique is the same regardless of what recursive relation $R$ you want.
