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

Related videos on Youtube

Zermie
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

Comments

  • Zermie
    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
      Git Gud over 8 years
      You can't prove it. See this.
    • Zermie
      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
    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
    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
    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
    Martín-Blas Pérez Pinilla over 8 years
    OK, stupid phrasing. "The axiom of regularity is not required".
  • Zermie
    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
    Andrés E. Caicedo over 8 years
    Sure, that is typically how we proceed. The separation axiom is particularly useful in practice.
  • Zermie
    Zermie over 8 years
    Great! Thank you for your help, Andres!
  • Admin
    Admin about 5 years
    @AndrésE.Caicedo, "If your class is a set, so is the collection of all sets". How so? Please can you step through why this is the case? Thanks!