Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have the following statements:

  • Definition of set. $y$ is a set if and only if there is $z$ such that $y\in z$; otherwise $y$ is a proper class
  • The comprehension axiom. Given a formula $\psi(x)$ where $x$ occurs free in $\psi$ and a class $y$, we have that $y\in\{x:\psi(x)\}$ if and only if $y$ is a set and $\psi(y)$
  • Definition of the empty set. $\emptyset$ is defined as $\{x:x\neq x\}$
  • Definition of the universe. $V$ is defined as $\{x:x=x\}$

At this point I managed to prove that:

  • no set belongs to $\emptyset$
  • every set belongs to $V$
  • $\{x:x\notin x\}$ is a proper class

How can I prove that $\emptyset$ is a set and that $V$ is a proper class? How can I prove that $\{x:x\notin x\}$ is equal to $V$?

Do I need additional statements perhaps?

Thank you.

share|improve this question

3 Answers 3

To show the empty set is a set use the power set axiom, as it appears in every power set.

To show the universe is not a set, again use the power axiom and deduce from that $V\in V$ to derive a contradiction.

Edit: We first prove some lemma, suppose $M$ is a proper class then it is non-empty.

Indeed, assume $M$ is an empty class, take any $x$ then $\{y\in x\mid y\in M\}$ is a subset of $x$, i.e. an element of the power set of $x$, which is the set of all subsets of $x$.

By extensionality we can now prove that this is the empty set.

The others two questions are of a similar state of mind, using the axiom of foundation if you prefer to avoid the power axiom.

share|improve this answer

To prove that $\emptyset$ is a set, use the Empty Set Axiom (if it's on your list).

To prove that $V$ is a proper class, obtain a contradiction if it's a set, using $\{x : x \notin x\}$.

Finally, the Axiom of Regularity should imply that $x \notin x$ for all sets.

share|improve this answer
    
Surely you don't need regularity for the last one? It's just Russell's paradox: if $y=\{x\mid x\notin x\}$ is a set, then either $y\in y$, in which case $y\notin y$, or $y\notin y$, in which case $y\in y$, a contradiction. –  Arturo Magidin May 17 '11 at 20:53
    
@Arturo: I want to prove $\forall x, x \notin x$. –  Yuval Filmus May 17 '11 at 21:15
    
Ah; but even in the absence of regularity, $\{x\mid x\notin x\}$ is still a proper class (even though it may not be equal to $\mathbf{V}$). –  Arturo Magidin May 18 '11 at 3:15

With the axioms you gave, it is impossible to prove that $\emptyset$ is a set, because your axioms are consistent with a model in which everything is a class and there are no sets at all.

Most set theories have an axiom that rules out this model. In ZF, there is an "axiom of infinity" which postulates the existence of a set $\mathbb N$ (with certain properties we do not need right now), and since $\mathbb N$ is a set, $\emptyset$ is an element of $2^{\mathbb N}$ and is therefore a set. In Morse-Kelley theory, the axiom just says that "there exists a set", which is enough to rule out the model in which there are no sets.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.