Is the axiom of existence sufficient to show that replacement implies comprehension?

Question

It is easy to show that if one accepts the axiom of the empty set (there exists a set for which no other set is its member) and the rest of the ZFC axioms one can show that the axiom scheme of replacement implies the axiom scheme of comprehension. What I was wondering is if it is possible to show the implication using the axiom of existence (there exists a set) instead of the empty set axiom. I was thinking about using the axiom of foundation and induction but that would require the class of ordinals which contains the empty set. Also since the definition of the axiom of infinity requires that the empty set is defined it too must be excluded.

Answer 1

No, you cannot prove Separation from ZFC without Separation, Empty Set, and Infinity (note that it is irrelevant to assume a set exists, since if no sets exist then Separation holds vacuously). Here is a sketch of how to construct a model that is a counterexample.

Start with an infinite sequence of sets $x_0=\{x_1\},x_1=\{x_2\},\dots$, and then let $M$ be the collection of all hereditarily finite-and-nonempty sets you can build from the $x_n$. (So for instance, a typical element of $M$ is a formal expression like $\{x_3,\{x_1,\{x_0,x_{16}\}\},\{x_1,x_5,x_8\}\}$. It is easy to see that $M$ satisfies all of ZFC except Separation, Empty Set, and Infinity. (For Regularity, note that although $M$ is not well-founded, the collection $\{x_0,x_1,x_2,\dots\}$ which would witness that is not a set in $M$ and every element of $M$ only contains finitely many of the $x_n$.) But clearly $M$ does not satisfy Separation since it has no empty set.

(Here's another way to describe this $M$. Define $f:V_\omega\to V_\omega$ recursively by $f(\emptyset)=\{\emptyset\}$ and $f(x)=\{f(y):y\in x\}$ if $x$ is nonempty. Then $M$ is the direct limit of the sequence $V_\omega\stackrel{f}\to V_\omega\stackrel{f}\to V_\omega\stackrel{f}\to\dots$, with $x_n$ being the image of $\emptyset$ from the $n$th copy of $V_\omega$.)

Note, though, that in the absence of Infinity, Regularity is rather weak, and is usually replaced by a stronger axiom scheme that says every nonempty class has an $\in$-minimal element. With this stronger version of Regularity (and the existence of a set), Empty Set follows immediately, since an $\in$-minimal element of the entire universe must be empty. So if you drop Separation, Empty Set, and Infinity, but replace Regularity with its more "proper" statement in the absence of Infinity, then you can still deduce Separation.

(Alternatively, if you choose to just drop Regularity entirely, there is an even easier counterexample: just take a model $M$ with only one element $x$ satisfying $x\in x$.)