# A set is not an element of itself.

2,762

## Solution 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 $A$.

## Solution 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.

Share:
2,762 Author by

### Zermie

I'm currently studying Mathematics at the university level in the US. I'll be graduating soon, but I would like to keep studying Maths as a hobby. This, in particular, is why I made this profile. That is, I can continue studying without having to be enrolled into an actual class. Just some of the privelidges of living in the 21st century, I suppose! FIAT LUX!

Updated on August 01, 2022

• Zermie 4 days

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

• Git Gud over 8 years
You can't prove it. See this.
• Zermie over 8 years
Haha well that would make sense as to why I couldn't get a contradiction. Thank you! I'll make an edit as to what my homework problem is.
• Andrés E. Caicedo over 8 years
Why is it too large? Showing that is the point of the exercise. What does your last sentence mean?
• Zermie over 8 years
Ah, thank you! This makes perfect sense. I knew it was something very simple and elegant that would do the trick. :)
• Zermie over 8 years
The axiom of regularity implies that a set is never an element of itself according to this, and as the above set would be an element of itself...
• Martín-Blas Pérez Pinilla over 8 years
OK, stupid phrasing. "The axiom of regularity is not required".
• Zermie over 8 years
Just a quick question: If we're trying to show that a class IS a set, is it sufficient to show that for at least one of these axioms we get a class that we know for sure is a set? E.g.: Taking the unionset and getting a class that we know to be a set.
• Andrés E. Caicedo over 8 years
Sure, that is typically how we proceed. The separation axiom is particularly useful in practice.
• Zermie over 8 years
Great! Thank you for your help, Andres!
• 