In the paper "Internal Set Theory: A New Approach to Nonstandard Analysis", Edward Nelson gives the following definition and claim during his proof of the conservativity of ZFC + IST over ZFC:
"We let $R(0) = \emptyset$ and define by transfinite induction for all ordinals we let \begin{align*} R(\alpha) = \bigcup_{\beta < \alpha} P(R(\beta)). \end{align*} Every set $x$ is in $R(\alpha)$ for some ordinal $\alpha$."
The broader context for this claim is that Nelson is building a sufficiently large universe such that given a formula $A$, its truth is equivalent to its truth when the free variables are bounded within that smaller universe.
While it is my understanding that all familiar mathematical objects can be built (at least in principle) out of the empty set, it is unclear to me how we rule out the existence of objects whose existence is not guaranteed by any of the axioms. Is this statement true?
For context, my background in ZFC is Paul Halmos' "Naive Set Theory", and my background in logic is Herbert Enderton's "An Introduction to Mathematical Logic".
Let $\text{WF} = \cup_{\alpha\in\text{Ord}} R(\alpha)$. Let $S$ be any set. By replacing $S$ by $$\operatorname{TrCl}(S) = \bigcup_{n\in\Bbb{N}} \left(\cup^n S\right)$$
we may assume $S$ is a transitive set. (A transitive set is a set such that $\cup S\subseteq S\subseteq P(S)$). Note that if $S\subseteq R(\gamma)$ then $S\in R(\gamma+1)$, so we need to prove there is some $\gamma$ for which $S\subseteq R(\gamma)$.
If for all $y\in S, y\in\text{WF}$, let $F(y)$ by the least ordinal $\tau$ for which $y\subseteq R(\tau)$. By an instance of the axiom of replacement, $\{F(y)+1:y\in S\}$ is a set, so there's an ordinal $\gamma=\sup\{F(y)+1:y\in S\}$. Since for each $y\in S$, $y\subseteq R(F(y))$ then $y\in R(F(y)+1)\subseteq R(\gamma)$. This proves $S\subseteq R(\gamma)$ which is what we wanted to show.
So finally, assume by contradiction that $S\setminus WF \ne \emptyset$. Since $S\setminus WF \subseteq S$, it is a set (by an instance of the axiom of comprehension.) So by the axiom of regularity, $S\setminus WF$ has an $\in$-minimal member. Let's call it $z$. If $p\in z$ then $p\in S$ because $z\in S$ and $S$ is transitive. Because of the $\in$-minimality of $z$ it follows that $p\not\in S\setminus WF$, so $p\in\text{WF}$. So all members of $z$ are in $\text{WF}$. By the same argument used in the previous paragraph, it follows that $z\in\text{WF}$ - in contradicition to the assumption that $z\in S\setminus\text{WF}$. This completes the proof.