How do I get the existence of a set in ZFC following Jech?

I am learning some set theory and logic on the side and am looking Jech's book, "Set Theory". At the moment, I am learning the basic axioms, and what I can and cannot do with them. Most of the axioms are of the form, if such and such set exists, then so does this set (the power set axiom, If X exists, so does it's power set). These sets by themselves do not give us actual sets to deal with, so the entire theory could be empty. Thus we have the axiom of infinity which states, $\exists S (\emptyset\in S \wedge (\forall x\in S)x\cup\{x\}\in S)$. However, this axiom seems to already give the existence of the empty set before one can get the infinite set, $S$. So do we need the empty set in existence before we can state this axiom or does it come from some other place. Again it does not seem to come from the other axioms.

• Apr 16 '12 at 6:54
• While it is perfectly reasonable to accept Brian's answers (since they are often excellent) it is much much better to wait a bit longer than 30 minutes, and surely longer than 8 minutes since the answer was posted. Apr 16 '12 at 7:07
• This issue is discussed to some extent here Apr 16 '12 at 7:56
• Meta thread related to Asaf's comment: Length of time to wait before accepting an answer Apr 16 '12 at 10:47
• @Martin Sleziak: if you simply replace $\emptyset \in S$ with its definition, so that Jech's axiom is actually a formula in the language of ZFC, it does imply that there is a set, because it starts with an existential quantifier. Apr 16 '12 at 11:05

The usual formalizations of first-order logic implicitly assume a non-empty domain of discourse. Knowing that, one can argue informally as follows: let $D$ be any set, and let $E=\{x\in D:x\ne x\}$. The existence of $E$ follows from comprehension, and it’s not hard to prove that $\forall x(x\notin B)$.

• So their is a hidden axiom: A set exists. Apr 16 '12 at 7:04
• When you said "follows from extensionality" I believe you meant "follows from separation". Extensionality is a universal axiom after all. Apr 16 '12 at 10:58
• @Carl: Actually, I meant comprehension. :-) Thanks. Apr 16 '12 at 17:24
• @Carl: (and Brian) I've seen "axiom of separation" and "axiom of restricted comprehension" used synonymously.
– user14972
Apr 16 '12 at 18:25
• @Hurkyl: Indeed. And axiom of comprehension is yet another synonym, and the one to which I’m most accustomed. Carl and I just come from different terminological traditions. Apr 16 '12 at 18:30

The symbol $\emptyset$ is not part of the formal language of ZFC, so from the strictest viewpoint the subformula "$\emptyset \in S$" is not even syntactically valid in ZFC.

One way to handle that problem is to prove there is a set that has no members, as Brian M. Scott indicates, and then make a definitional expansion of ZFC to add a constant symbol $\emptyset$ for this set.

Another way to handle it, without making a definitional expansion, is to just mentally replace "$\emptyset \in S$" with something like "$(\exists z)[(\forall w)[\lnot (w \in z)] \land z \in S]$". Obviously this makes the axiom much harder to read, so writing $\emptyset \in S$ is a convenient abbreviation.

This same issue comes up in many formal settings: when someone uses a symbol that is not in the formal language, but where you know the intended definition for the symbol, you can simply interpret the formula as an abbreviation for a longer formula that does not use the symbol. For example, the definition in $\{z \in \mathbb{N} : z\text{ is even}\}$ is not a formula of ZFC, but it is an abbreviation for $(\exists w \in \mathbb{N})[z = w + w]$, which in turn is an abbreviation for a much longer formula that does not include the symbols "+" or "$\mathbb{N}$". In most settings the author will not comment much on this sort of thing unless it is unclear that there is a definition in the language of set theory or unless it matters which specific definition is used.

You can replace $\emptyset$ with $\{ x\in S: x\neq x\}$ in the axiom of infinity, because of the separation axiom. This way, the axiom of infinity implies the existence of all sets all by itself.

• +1 Nice solution. In words: there exists $S$ such that its subset $E$ of elements unequal to themselves satisfies $E\in S$ and [rest of requirements for $S$]. Alternatively one could replace $\emptyset\in S$ in the condition for $S$ by $\exists E\in S(x\in E\implies x\neq x)$; in words $S$ contains an element $E$ with the property of being empty. Apr 16 '12 at 8:03
• I would take this as the correct answer. The existence of a set follows from the axiom of infinity, even using a modification of first order logic that would allow for empty domains. Apr 16 '12 at 9:01
• in ZFC, the {} set-builder notation is not part of the formal language of the theory. In particular it does not give actual terms in the sense of first-order logic, and so you cannot substitute it into axioms in actual formal derivations. From the viewpoint of ZFC the set-builder notation is just shorthand we use in the metatheory. Apr 16 '12 at 10:43