# 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.

• The most general 'formal justification' takes the form of an appropriate recursion theorem, but in this case replacement gives $V_\lambda$ simply by observing that $\lambda$ is a set, $V_\beta$ for all $\beta<\lambda$ is a set, and $\langle V_\beta:\beta<\alpha\rangle$ is a function whose domain is $\lambda$ with the image $V_\beta$ of each element $\beta\in\lambda$ a set, so the range $\{V_\beta\}_{\beta<\lambda}$ of this function is also a set -- the union of this range is $V_\lambda$, as you suggest. Jan 6, 2023 at 5:01
• Typo: $\alpha$ should be $\lambda$ in the above comment. Jan 6, 2023 at 6:41
• @AlecRhea But wouldn't the domain of that function be $V_\lambda$ itself? Unless it's a class function, in which case my comments apply. Jan 6, 2023 at 6:45
• No, the domain is $\lambda$ -- it is the function sending each ordinal $\beta<\lambda$ to the chunk of the cumulative hierarchy up to $\beta$. As long as each of these initial chunks of the hierarchy at each successor step are defined, the function is well-defined. Replacement then gives that its range is a set, and the union of this range is $V_\lambda$. Jan 6, 2023 at 6:58
• @AlecRhea My apologies, I mean the codomain. Jan 6, 2023 at 7:06

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 any 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 outline 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 $$V_\lambda$$.

• This is a very enlightening answer, thank you! Jan 6, 2023 at 8:41
• @ViHdzP Glad to help! Jan 6, 2023 at 8:42

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.