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As you know, set is an undefined concept in ZFC. But set theorists often talk about something being a set while other things are not sets. For example, $\{x|x\notin x\}$ is not a set but $\{x\in S|\phi(x)\}$ is a set given $S$ is a set.

So, what does the sentence "$S$ is a set" actually mean?

J W
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peter
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  • "As you know, set is an undefined concept in ZFC" - I didn't know that and it seems to be not true. Why do you say this? Sets are elements of ZFC models (the things it is made up of), by definition. Why do you think that this is not a definition? – Martin Brandenburg Dec 08 '24 at 12:30
  • @MartinBrandenburg I say this because I see many people say this, e.g., https://math.stackexchange.com/questions/1452425/what-is-the-definition-of-a-set. – peter Dec 08 '24 at 12:56
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    @MartinBrandenburg I don't find that definition coherent. You could turn the question around and ask: what is not a set? I don't think you can find such an item, because any item can be an element of a model of ZFC. All that is required to be a model of ZFC is a collection $V$ of items together with an appropriate binary relation $\in$. So anything that you think is not a set, call it $x$, can be made a set by picking a ZFC model $V$ and an element $y \in V$, then define $V' = (V \setminus {y}) \cup {x}$, and translate the $\in$-structure on $V$ across the obvious bijection $V \to V'$. – user43208 Dec 08 '24 at 13:48
  • @MartinBrandenburg: I definitely think that defining "sets" in terms of ZFC models is putting the cart before the horse. In order to make sense of the assertion that $(M,\varepsilon)$ is a model of ZFC, you already need to be working in a metatheory that allows you to reason about sets (usually that metatheory is ZFC itself). – Joe Dec 24 '24 at 22:33

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When we discuss about set theories in which everything is a set (e.g., $\sf ZFC$ and its close relatives, but also $\sf NF$ and $\sf GPK$), for something to be a set means that something exists.

More often than not, we use this when we refer to an extensional definition, e.g. "$\{x\mid x\in x\}$ is a set" or "$\{x\mid x\notin x\}$ is not a set". And more broadly, if $\varphi(x)$ is a formula, $\{x\mid\varphi(x)\}$ is or isn't a set.

What we really mean when we say that $\varphi$ defines a set, or "$\{x\mid\varphi(x)\}$ is a set", is $$\exists X\forall x(x\in X\leftrightarrow\varphi(x)).$$ Or, in other words, there is a set (or simply an object in our universe) which captures exactly all the things that have property $\varphi$.

Originally suggested by Frege, the Comprehension axiom stated that every property defines a set. But people had started poking holes in that concept, until it was fully laid to rest by Russell in his now-famous paradox: $x\notin x$ does not define a set.

To overcome this, set theories generally find some kind of way to restrict the unwieldy power of Comprehension.

  • Zermelo did it by limiting $\varphi$ to formulas of the form $x\in S\land\varphi(x)$, or in a more familiar way: given a set $S$, $\{x\in S\mid\varphi(x)\}$ is a set;
  • Quine did it in $\sf NF$ by requiring that $\varphi$ is stratified (the variables have a "typing" to them, so $x\in x$ and $x\notin x$ are not valid formulas anymore);
  • Esser did it in $\sf GPK$ by requiring that the formula does not use the negation symbol, so $x\notin x$ is not a valid formula for Comprehension.

Some additional care is needed, though. In all of these set theories we can come up with formulas which "technically break the rules" but still define sets. For example, $x\neq x\land\exists y(y\in x\land x\in y)$ would not be permissible in $\sf ZFC$ since it is not bounded by any set $S$, the formula is also not positive, and cannot be stratified. Nevertheless, the formula clearly defines the empty set, which exists in all of these set theories.

Asaf Karagila
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There are various "flavors" of set theory.

If we restrict our consideration to ZF (without urelemets) where everything is a set, we can rely on standard semantics for Classical Logic where "an interpretation for the language L of the theory is a structure with a non-empty domain-of-discourse, or simply domain".

In this case, we may consider a term $a$ that denotes the existing object of the domain and thus $∃x(x=a)$ holds, that reads: "there is an object in the domain of sets".

If instead we are working with set theory with classes, like NBG , we have that everything is a class and a set is a class that belongs to another class.

Thus, we introduce a suitable predicate $\text M(x)$ (from the German Menge): $\text M(x)↔∃y(x∈y)$.

See also What is the definition of a set.

Mauro ALLEGRANZA
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