Pure set theory, wherein all objects considered are sets —whose elements are themselves sets, and so forth— is usually thought of as building itself up in an ex nihilo fashion off the empty set $\emptyset$: one starts with $\emptyset$, then constructs off it a variety of sets like $\{\emptyset\}$, $\{\{\emptyset\}\}$, $\{\emptyset, \{\emptyset\}\}$, etc., which can later be interpreted in various forms (von Neumann ordinals, naturals, reals via Dedekind cuts, etc.) to make up the (near) totality of all everyday mathematical objects (groups, rings, functions, etc.). My question is: it seems intuitive that under one such pure set theory, all objects considered are 'built off' $\emptyset$, so that if one were to 'look into' some set long enough, one would eventually find a $\emptyset$ off which all other elements were constructed.
Now, for this to be the case, it seems intuitive to me that such a pure set theory would need something akin to Regularity, so that infinitely regressive chains of inclusion are prohibited and our 'looking into' the foundations of some set must eventually come to an end. From what I've read, Zermelo set theory $\mathsf{Z}$, i.e., Zermelo-Fraenkel without Replacement ($\mathsf{ZF} – \mathsf{R}$), is unable to prove the existence of the image-sets containing all power-sets (or union-sets, or singletons) of a particular set $X$. That is, under $\mathsf{Z}$, the sentence $\forall X \ \exists Y \ \forall z \ (z \in Y \leftrightarrow \exists n \in \omega \ \phi(n, z))$, where $\phi(n, z)$ intuitively expresses the (outside) notion that '$z = \mathcal{P}^n(X)$,' is undecidable. This seems to be the case also for the possibility of a general formal notion of $\bigcup^nX$ ($n$-union) and $\underbrace{\{\dots\{X\}\dots\}}_{n \text{ times}}$ ($n$-singleton) for every $n \in \omega$, and hence motivates the inclusion of Replacement.
Hence I ask: Given some pure set theory at least as strong as $\mathsf{Z} + \mathsf{R}$ (or perhaps of an equivalent strength/able to deduce the existence of the image-set of all $\bigcup^nX$ unions for all $X$ and $n \in \omega$) with something akin to Regularity, is it the case that: $\forall X \ \exists n \in \omega \ \emptyset \in \bigcup^n X$?
Or perhaps one can prove (metatheoretically) that: for every $X$ in the theory, there is some $n \in \mathbb{N}$ such that $T \vdash \emptyset \in \bigcup^n X$, where $n$ is an (outside) numeral off which the finite construction $\bigcup^n X = \underbrace{\bigcup \dots \bigcup X}_{n \text{ times}}$ is made and $T$ is the theory under consideration?
Finally, perhaps the statement should be extended to all ordinals (not just the finite) to work? Does the (again, outside/informal/metatheoretic) statement: for every $X$ in the theory, there is some $\alpha \in \mathrm{Ord}$ such that $T \vdash \emptyset \in \bigcup^\alpha X$ provide some more significant import?
Best Answer
"Looking into a set long enough" means the same thing as looking into its transitive closure. As you indicate that's the same thing as iterating the union $\omega$ times. In other words, $\operatorname{trcl}(X) = \bigcup_{n\in\omega} \bigcup^nX.$ (You don't get anything else by iterating more, since the RHS is already transitive, so $\bigcup$ doesn't give you anything that isn't already inside).
Yes, we can always find the empty set in the transitive closure of a nonempty set. If not, let $x$ be a $\in$-minimal element of $\operatorname{trcl}(X)$. If $x\ne\emptyset,$ there is some $y\in x,$ but by transitivity, $y\in \operatorname{trcl}(X),$ contradicting the $\in$-minimality of $x.$ So actually, we've shown something a little stronger: transitive sets / transitive closures always have $\emptyset$ as a the only $\in$-minimal element.
And yes, regularity is crucial here, as a Quine atom is a clear counterexample.
As you alluded to, an instance of replacement was used in the above to show that the transitive closure of a set is a set. The axiom of infinity was also used for this purpose. They are both strictly necessary to show that every nonempty set's transitive closure contains the empty set$.^*$
But their role is pretty minor in the sense that if you upgrade regularity to an $\in$-induction schema (i.e. every nonempty class has an $\in$-minimal element) then you no longer need them... so in that sense, regularity is what's really doing the work here, as you might expect.
$^*$In both the case of dropping infinity and dropping replacement there is a countermodel that essentially adds distinct elements $x_0,x_1,x_2,\ldots$ with $x_n = \{x_{n+1}\}$ to an appropriate initial segment of the cumulative hierarchy such that the model has the sequence as a class, but only has finitely many $x_n$ in any set. Because of this, any set has an $\in$-minimal element, so regularity holds. However, the class of all the $x_n$ clearly has no $\in$-minimal element, and the transitive closure of $x_n$ is $\{x_{n+1},x_{n+2},\ldots\}$, which is not a set and doesn't contain the empty set.
In more detail, work in ZF - regularity + "there is a sequence $(x_n: n\in \omega)$ such that for all $n$, $x_n = \{x_{n+1}\}$", whose consistency can be established e.g. by a Rieger-Bernays permutation model. Let $M_0=\emptyset$ (in the case where we are omitting the axiom of infinity) or $M_0=V_\omega$ (in the case where we are omitting replacement). Then recursively let $M_{n+1} = M_n \cup\{x_n\} \cup P(M_n),$ and $M = \bigcup_n M_n.$ Then $M$ is a transitive model of ZF-infinity (in the first case) or ZF-replacement (in the second) that fits the above description.