I've been trying to work out this proof, but I think there's something I'm missing out because with the deductions I made, it much looks like the universal class is a set... (help).
So in my class we're using Von Neumann-Bernays-Gödel's axiomatic set theory (NBG), which is also presented in Dugundji's "Topology" (1966). The definitions, propositions and axioms I have to work until now to demonstrate the stated are the following:
Definition $\left( A \subset B \right) \iff \left( \forall x : x \in A \implies x \in B \right)$
Definition $\left( A = B \right) \iff \left( A \subset B \right) \wedge \left( B \subset A \right)$
Definition (of a Set) The class $x$ is a set only if there exists a class $A$ such as $x \in A$.
Proposition For all class $x$, $x=x$.
Proposition $\left( x \in A \right) \wedge \left( A = B \right) \implies \left( x \in B \right)$
Axiom I (of Individuality) $\left(x \in A\right) \wedge \left(x = y\right) \implies y \in A$
Axiom II (of Class Formation) $\left( x \in A \right) \iff \left( x \,\,\text{is a set} \right) \wedge p(x)$.
And the universal class was defined as $U := \{ x : \left( x \,\,\text{is a set} \right) \wedge \left( x = x \right) \}$. Also, a proper class was defined as every class that is not a set.
The exercise is "Prove that the universal class $U$ is a proper class".
So the way I'm approaching the demonstration goes like this:
We want to prove that the universal class $U$ is a proper class $\iff$ we want to prove that the universal class $U$ is not a set.
By contradiction, let's assume $U$ is a set. In general, $U$ is still a class so it satisfies $U = U$. Then, by the definition of $U$ it follows that $U \in U$ (because it is a set and it is equal to itself).
...And that's everything I have! I haven't been able to find any contradictions to this procedure. Also, I haven't come up with any other ideas to approach the exercise.
I would really appreciate your help!
P.S. I tagged this with "analysis" because it's the course I'm learning this topic in!
I think you probably need Axiom of Sifting, which states that if $A$ is a set and $\mathscr A$ is a class, then $A\cap \mathscr A$ is a set. This axiom is stated below axiom of class formation but before the problem "8.10 The class of all sets is not a set" in the book. It has a similar function of Axiom of Comprehension in ZFC.
If this is allowed, then the proof goes as follows: assume $\mathscr U$ is a set, then the Russell class $\mathscr R$, defined through Axiom II as $$x\in\mathscr R\iff x\text{ is a set}\;\wedge\;\neg (x\in x)$$ is set. Indeed, Axiom of Sifting applied to $\mathscr R = \mathscr U\cap \mathscr R$ gives this result.
Assume $\mathscr R\in\mathscr R$. Then, by definition, $\mathscr R\notin \mathscr R$, contradiction.
Assume $\mathscr R\notin \mathscr R$. Since $\mathscr R$ (on LHS) is a set, and $\mathscr R$ (on RHS) contains all sets $x$ with $x\notin x$, $\mathscr R\in \mathscr R$. This is again a contradiction.
Without sifting, it is consistent with Axioms I and II to have the set of all set being a set. In fact, consider the following universe $U=\{\circ,\bullet\}$, where we define the binary relation $\in$ as follows:
Then, $\circ$ is a proper class, since no classes contains it. On the other hand, $\bullet$ is a set, which is coincidentally the set of all sets. We check that Axiom II holds for this universe:
Let $p(x)$ be a proposition with free variable $x$.
Case 1: $p(\bullet)$ holds. Then, $x\in \bullet\iff x\text{ is a set}\wedge p(x)$. So $p$ forms the class $\bullet$.
Case 2: $p(\bullet)$ does not hold. Then, $x\in \circ\iff x\text{ is a set}\wedge p(x)$. So $p$ forms the class $\circ$.
As you can see, $(U,\in)$ provides an example of axioms I and II. Therefore, it is not possible to exclude the set of all set from these axioms alone. Note $U$ clearly does not satisfy sifting: $\circ\cap\bullet=\circ$ is not a set.