My students are writing an essay on axiomatic set theory focusing on the concept of collectivizing predicates. The axioms in their work establish which predicates are accepted as collectivizing predicates (those that define sets) and serve as the basis for constructing the other sets in the theory. They have chosen the ZF axioms presented in the book "Introduction to Set Theory" by Hrbacek and Jech, which include the Axiom of Existence, Axiom of Extensionality, Axiomatic Scheme of Comprehension, Axiom of Pairing, Axiom of Union, Axiom of Power Set, and Axiom of Infinity.\

It is worth mentioning that my students do not work with the concept of class, and they have not yet explored the Axiom of Replacement or the Axiom of Regularity (Foundation). In addition, they have developed a calculatory style of argumentation, inspired by "Axiomatic Set Theory à la Dijkstra and Scholten" by Ernesto Acosta, Bernarda Aldana, Jaime Bohórquez, and Camilo Rocha, published in Advances in Computing, pp. 775-791, Springer 2017.\

Part of their work involves studying properties of collectivizing predicates, such as whether $\phi$ and $\psi$ being collectivizing, $\phi\wedge \psi$, $\phi\vee \psi$, $\phi\Rightarrow\psi$, and $\neg \phi$ are also collectivizing. This has led to the question: could it be true that $\phi$ is non-collectivizing if and only if $\neg \phi$ is collectivizing? One of them proposed studying the predicate $\emptyset \in x$. After much consideration, we decided to post this question on Stack Exchange.

In terms of proper classes the question is if complements of proper classes could be sets in ZFC. In particular:

Are {$x\mid \varnothing \in x$} and {$x\mid \varnothing \notin x$} proper classes inZF?

Now we understand (thanks to those who helped us) that the following argument: "Let A be the class {$x\mid \varnothing \in x$} and let R be the Russell class {$x\mid x \notin x$}. Since R belong to A then A must be a proper class." does't have sense.

  • $\begingroup$ I am pretty sure that the trial to construct the set of all sets containing the empty set or the set of all sets not containing the empty set leads to a very similar issue we run into , if we try to construct the set of all sets. $\endgroup$
    – Peter
    May 12 at 15:11
  • $\begingroup$ A proper class is too large to be an element of anything and in particular you cannot have a class of proper classes. So it may be difficult to say "$R$ belongs to $A$" $\endgroup$
    – Henry
    May 12 at 15:24
  • $\begingroup$ ... and when you say "Let $A$ be the class $\{x\mid \varnothing \in x\}$" the $x$s are sets not proper classes $\endgroup$
    – Henry
    May 12 at 15:33
  • $\begingroup$ What do you mean by " R belongs to A" ? A is a class which contains sets, and R is not a set. So, how can R belong to A? $\endgroup$ May 12 at 19:04
  • $\begingroup$ A non-zero von Neumann ordinal is bigger than zero, which is to say it contains $\emptyset$. The class of ordinals bigger than some fixed ordinal is already proper. $\endgroup$
    – Nikolaj-K
    May 16 at 0:02

4 Answers 4


In $\mathsf{ZF}$, for any set $s$, both $\mathcal A_s=\{x:s\in x\}$ and $\mathcal A_{-s}=\{x:s\notin x\}$ are proper classes. One way of seeing this is that a class $\mathcal C$ is a set if and only if it is bounded (it has a rank), that is, for some ordinal $\alpha$, $\mathcal C\subseteq V_\alpha$, but one easily checks that both $\mathcal A_s$ and $\mathcal A_{-s}$ have members of arbitrarily large rank.

Another argument is obtained by recalling that if $A$ is a set and the class $F$ is a function, then $F''A=\{F(a):a\in A\}$ is also a set. Note that any set $x$ in $\mathcal A_s$ can be written in a unique way as $y\cup\{s\}$, where $y\in \mathcal A_{-s}$. This means that, if $\mathcal A_{-s}$ is a set, then so is $\mathcal A_s$, but this would give us that $V=\mathcal A_s\cup \mathcal A_{-s}$ is a set, which it is not.

(Note, by the way, that in $\mathsf{ZF}$ the members of classes are sets, so we cannot argue about whether $\mathcal A_\emptyset$ is a set by noting that the Russell class would be a member of it. It is not, because the members of $\mathcal A_\emptyset$ are sets, by design if you wish, and the Russell class itself is not a set. But this is irrelevant to the issue. In other set theories we can have proper classes belonging to classes, and then we would have to argue differently.)

  • 1
    $\begingroup$ Thank you very much for your help $\endgroup$ May 15 at 14:27

I have another proof which uses the fact that in $\mathsf{ZFC}$ the union of a family of sets is a set provided this family forms a set, which is actually an axiom of $\mathsf{ZFC}$ called the axiom of union.

Lemma. The class of all sets $V=\{x | x=x\}$ is proper.

Proof. If $V$ is a set, then, since $x\not\in x$ is a well-formed formula of $\mathsf{ZFC}$ which has one free variable $x$, by the axiom schema of separation the class $\{x\in V | x\not\in x\}$ is also a set, which is equal to the Russell class, a contradiction.

Theorem. The class $A=\{x|\varnothing\in x\}$ and the class $V\setminus A=\{x|\varnothing\not\in x\}$ are both proper classes.

Proof. If $A$ is a set, then so is its union $$\bigcup_{x\in A}x$$ by the axiom of union. But for any set $y$ we have $\{\varnothing,y\}\in A$, so $y$ is in the union. So union is equal to $V$ and this is a contradiction by the lemma.

Similarly, if $V\setminus A$ is a set, then its union $$\bigcup_{x\in V\setminus A}x$$ is a set. This union is equal to the class $V\setminus\{\varnothing\}$ since we have $\{y\}\in V\setminus A$ for any non-empty set $y$. Again, by the axiom of union, the union $$\bigcup_{y\in V\setminus\{\varnothing\}}y$$ of all non-empty sets is a set, and we can similarly argue that this union is equal to the proper class $V$. This is a contradiction, hence the class $V\setminus A$ is not a set.

  • $\begingroup$ Thank you very much. Very simple $\endgroup$ May 15 at 14:31

Theorem 1: If $S$ is a set, then $S$ does not contain every set $T$ such that $\emptyset \in T$.

Proof: Let $T = \{S, \emptyset\}$. By inspection, $\emptyset \in T$. Now consider the set $U = \{S, T\}$. By the axiom of regularity, $U$ must have an element which is disjoint from $U$. However, $T$ is not disjoint from $U$ (because $S \in T$ and $S \in U$), so $S$ must be disjoint from $U$. Since $T \in U$ and $S$ is disjoint from $U$, we know that $S$ does not contain $T$.

Theorem 2: If $S$ is a set, then $S$ does not contain every set $T$ such that $\emptyset \notin T$.

Proof: If $S$ is empty, the conclusion follows immediately, so we need only consider the case where $S$ is not empty. Let $T = \{S\}$. By inspection, $\emptyset \notin T$. Now consider the set $U = \{S, T\}$. The proof continues as above.


Notice that

for any formula φ(x) in the language of set theory:

V = { x | φ(x)} ∪ { x | ¬ φ(x) }

Where V is the Universe, the Proper class containing every set.

Since the union of two sets is a set, at least one of:

A = { x | φ(x)}

B = { x | ¬ φ(x) }

Must be a Proper Class.

In your case we have φ(x) = ∅∈x


A = { x | ∅∈x }

B = { x | ¬ ∅∈x }

Assume A is a set.

Then ⋃A is a set by the Union Axiom.

But ⋃A = V which is a proper class

Since, for any set x

{x} ⋃ {∅} is in A

so, every set is in the Union of A

Let's consider if B is a set. Notice that B is nonempty.

So b = {B} is a set ( by pairing axiom) and b∈B

So, {B}∈B∈{B}..... which Contradicts foundation.

To see why, Consider the set C = {b,B}

By Foundation all nonempty sets must be disjoint from one of their elements.

But, b∈C ∩ B and B∈C ∩ b a contradiction.

So, neither A nor B are sets in ZF


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