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I recently began reading, "The Foundations of Mathematics" by Kenneth Kunen and am having some difficulty understanding the rationale behind one of the very early proofs. The proof is about demonstrating the uniqueness of the empty set using the following 3 Axioms that Kunen establishes early in the book:

A. Set Existence: $\exists x (x=x)$ ...which Kunen will refer to as Axiom $0$

B. Extensionality: $\forall x,y \ \big( \forall z (z \in x \leftrightarrow z \in y) \rightarrow x=y\big )$

C. Comprehension Schema: $\forall z \big ( \exists y \forall x (x \in y \leftrightarrow x\in z \land \varphi(x) \big)$

I've seen the proof referenced many times so I am conceptually familiar with it...but I am having some confusion regarding the formal implementation of it. Kunen's proof is as follows:

(1) $\text{emp}(x):=\forall z (z\notin x)$

(2) By Extensionality, $\text{emp}(x) \land \text{emp}(y) \rightarrow x=y$

(3) To prove that $\exists y [\text{emp}(y)]$, start with any set $z$ (there is one by Axiom $0$) and apply comprehension with $\varphi (x)$ a statement that is always false (for example, $x \neq x$) to get a such that $\forall x (x \in y \leftrightarrow \text{FALSE})$ - i.e. $\forall x (x \notin y)$.

Combining (2) with (3) demonstrates that the empty set $\emptyset$ is unique.

I have several questions:

Firstly, how can we claim that $x \neq x$ is always false. By Set Existence, we claim that there is at least one set with the property that it is equal to itself...i.e. the relation "$=(x,x)$" holds true. In order to claim that $x \neq x$ is always false for any object $x$, shouldn't our Set Existence axiom be of the form $\forall x (x=x)$?

Secondly, given that Set Existence is presumably written correctly, what exactly is its purpose? Kunen's remark in (3), "...there is one by Axiom $0$..." seems to suggest that this axiom asserts that our domain of discourse is non-empty. But what exactly does that mean? My interpretation of a statement like that is, "Ok...so we know that we have at least one object in our domain of discourse, but no comment is made regarding whether we have more than one". Assuming that interpretation is correct, how is it that I can even generate multiple empty sets (say $\emptyset^*$ and $\emptyset '$) in (3) when the claim of the Comprehension Schema is that, "For any given set $z$ I can construct at least one set $y$ such that..."?

Said differently, if I only know for sure that one object $z$ exists (namely the $z$ that would be guaranteed to exist by Set Existence), shouldn't I only be able to generate a single empty set from it...i.e. because the Comprehension Schema states that I can construct at least one subset...but makes no comment about whether or not a can construct more than one ($\emptyset^*$ would be the first instance of an empty set construction but how could I also be guaranteed to construct the other $\emptyset'$?). The only thing I could think to do is use a different $\varphi$ (call it $\psi$) that also always evaluates to FALSE but I am unsure if that is the correct route.

Hopefully this rambling was expressed in a comprehensible manner. Any insight is greatly appreciated!

S.C.
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    The purpose of Axiom $0$ is to ensure that there is at least one set; the predicate $x=x$ is used simply because it is always true, just as $x\ne x$ is used in the proof because it is always false. In effect the axiom just says there is a set. That $x=x$ is always true (and hence that $x\ne x$ is always false) comes from the axioms of equality. You cannot generate a different empty set: extensionality implies that if $x$ and $y$ both have no elements, then $x=y$. – Brian M. Scott Oct 29 '20 at 23:46
  • @BrianM.Scott hmmm. I'm not sure how to ask this question...so please let me know if it's gibberish and I'll try to make an adjustment. Kunen has made no mention of the Axioms of Equality. SO, where exactly do the axioms of equality "rank" in relation to set theory axioms? Your comment seems to suggest that the notion of axioms of equality are actually established BEFORE set theory. In a sense, it seems like the axioms of equality are actually axioms about first order logic...in particular, axioms about how the binary relation 'equals' behaves in an FOL. Is that the right idea? – S.C. Oct 29 '20 at 23:52
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    Yes. I don’t know that particular book, but I suspect that Ken is assuming first-order logic with equality as the formal system in the background. That is both the more common approach and the one that he took in his graduate set theory course. – Brian M. Scott Oct 30 '20 at 00:04
  • Oh, great! That clears up the first question and the first part of the second question, but I am still uncertain why I can claim that both $\emptyset^*$ and $\emptyset'$ exist...unless of course I am correct that I need to create a different predicate $\psi$ that also always evaluate to false. – S.C. Oct 30 '20 at 00:15
  • I’m not quite sure what you’re asking: if you have another predicate $\varphi(x)$ that is always false, you can use it to define the empty set, but it will be the same empty set, because extensionality guarantees that there cannot be more than one. – Brian M. Scott Oct 30 '20 at 00:19
  • Let me clarify. I am trying to understand the step that is required BEFORE invoking extensionality. In the formulation of extensionality, we have references to two variables $x$ and $y$ (i.e. $\forall x,y [\forall z (z \in x \leftrightarrow z \in y) \rightarrow x=y]$). What I am trying to do is instantiate $x = \emptyset^$ and $y=\emptyset'$ so that I can ultimately conclude, via extensionality, $\emptyset^ = \emptyset'$ – S.C. Oct 30 '20 at 00:22
  • But that’s unnecessary: $\forall z(z\notin x)\land\forall z(z\notin y)$ implies $\forall z(z\in x\leftrightarrow z\in y)$, and $x=y$ then follows from extensionality. – Brian M. Scott Oct 30 '20 at 00:26
  • Thanks for keeping up with me this long! But I'm not sure I understand the rationale here. I interpret your conjunction as saying, "If you produce two sets for me, neither of which contains any elements, then the two sets must be equal". When I see that, I think to myself, "Well, I guess I need to construct (or demonstrate the existence of) two such sets." My attempt at doing this was to create $\emptyset^*$ and $\emptyset'$. But you said this is unnecessary...and I'm uncertain why. – S.C. Oct 30 '20 at 00:32
  • Extensionality implies that if $x$ and $y$ are both empty, then they are the same set; you don’t have to produce any empty set in order to prove this. In other words, it proves that there is at most one empty set. Axiom $0$ and comprehension give you an empty set, so it must be the empty set. – Brian M. Scott Oct 30 '20 at 00:37
  • I think I understand what you are trying to convey (sorry for confusion on my end): "I have just made a symbolic argument that if you construct two objects with these properties, then the two objects will ultimately be equal to one another. Therefore, you only need to construct a singular object with said property because we know that any other object constructed to possess this property will ULTIMATELY be exactly equal to our original object". Is that right? – S.C. Oct 30 '20 at 00:40
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    Essentially, though I’d omit ultimately and say simply that it must necessarily be equal to the original object. Ultimately implies a temporal sequence where none really exists. – Brian M. Scott Oct 30 '20 at 00:46
  • Thanks for all of the help! Greatly appreciated. – S.C. Oct 30 '20 at 00:47
  • You’re welcome! – Brian M. Scott Oct 30 '20 at 00:48
  • $x \ne x$ is always false because it is the negation of FOL equality axiom $x=x$ and every axiom/theorem of FOL is valid, i.e. always true.
    • – Mauro ALLEGRANZA Oct 30 '20 at 07:39
  • For poof of uniqueness of $\emptyset$, see this very similar post – Mauro ALLEGRANZA Oct 30 '20 at 07:40
  • IF we agree with the standard convention of semantics for FOL that every interpretation must have a non-empty domain, then YES $\exists x (x=x)$ is a consequence of equality axiom $\forall x (x=x)$, and thus it is superfluous.
    • – Mauro ALLEGRANZA Oct 30 '20 at 07:42