A set is not an element of itself.

Question

I know that in modern set theory that for a given set $A$, $A \notin A$, specifically by the axiom of regularity. However, I'm not permitted to use this axiom in my proof. What I am permitted to use are the following:

1. Axiom of Extensionality
2. Emptyset and Pairset Axiom
3. Separation Axiom
4. Powerset Axiom
5. Unionset Axiom
6. Axiom of Infinity

Edit: Thanks to @Git_Gud, I now am aware that a proof of this isn't possible.

The problem that I'm working on is this:

Determine whether the following class is a set or not:

$$\{x | x \text{ is a nonempty set} \}$$

I know this isn't a set by the axiom of regularity. However, I'm not allowed to use this as it appears later on in my textbook. The textbook is, "Notes on Set Theory" 2nd ed., by Moschovakis.

Any tips are appreciated! Thank you! :3

Answer 1

If your class is a set, so is the collection of all sets, which I assume you have already shown is not the case. To see this, note that $\emptyset$ exists, and therefore so does $\{\emptyset\}$ and $A\cup\{\emptyset\}$ for any set $A$.

Answer 2

{x|x is a nonempty set} is a proper class (not a set) because is too large (is the universe minus $\{\emptyset\}$). And, as I know, the axiom of regularity is not required for this.

Details of "is too large":

If $A=\{x|x\ne\emptyset\}=V-\{\emptyset\}$ set then by union axiom $V$ is a set. If $V$ is a set, by powerset axiom $P(V)$ is a set. But $P(V)=V$, impossible by Cantor.

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