Prove that $\mathfrak{U} \setminus (\mathfrak{U} \times \mathfrak{U})$ is not a set Let $\mathfrak{U}$ denote the universe we are working in and $\mathfrak{U} \times \mathfrak{U}$ is the class of all ordered pairs $z = (x, y)$ where $x, y \in \mathfrak{U}$. 
I know that neither $\mathfrak{U}$ nor $\mathfrak{U} \times \mathfrak{U}$ are sets. However, I am not sure how to go about showing that $\mathfrak{U} \setminus (\mathfrak{U} \times \mathfrak{U})$ is not either. 
The definition of a set I am using is the following:

A class $X$ is called a set if there exists a class $Y$ such that $X \in Y$, i.e. $X$ is a set $\iff (\exists Y) X \in Y$. 

So, for this problem, I need to show that there does not / cannot exist a class $Y$ with $\mathfrak{U} \setminus (\mathfrak{U} \times \mathfrak{U})$ as a member. I figured that I should try and assume that such a class $Y$ exists and then (hopefully) reach a contradiction, but I am not sure how to proceed. 
Edit: Part of my issue is that I am having difficulty conceptualizing the class $\mathfrak{U} \setminus (\mathfrak{U} \times \mathfrak{U})$. I understand that it means "the universe without ordered pairs"... so does that mean it is a subclass of $\mathfrak{U}$? 
 A: Consider the class $S':=\big\{x\cup\{x\}:x\in U\big\}$.
Fix $x\in U$. Suppose $x\cup\{x\}=(y,z)$ for some $y,z\in U$. Then
$$x\cup\{x\}=\{\{y\},\{y,z\}\}.$$
Therefore $x=\{y\}$ or $x=\{y,z\}$.
Because $y\in x$, it follows that $y=\{y\}$ or $y=\{y,z\}$.
This contradicts the axiom of regularity---sets cannot contain themselves.
As a result, $S'\cap (U\times U)=\varnothing$.
We have $S'\subseteq U\setminus(U\times U)$.
Observe $S'$ is not a set. Indeed, $\bigcup S'=U$. The desired result follows---$U\setminus(U\times U)$ is not a set.
A: Every element of $\mathfrak{U} \times \mathfrak{U}$ has cardinality equal to either one or two.  Therefore, for the proper class $\mathfrak{O}$ of von Neumann ordinals, we have
$$\mathfrak{O} = \{ 1, 2 \} \cup [\mathfrak{O} \cap (\mathfrak{U} \setminus (\mathfrak{U} \times \mathfrak{U}))].$$
Now, if $\mathfrak{U} \setminus (\mathfrak{U} \times \mathfrak{U})$ were a set, then the right hand side would also have to be a set.  (For one step, $\mathfrak{O} \cap (\mathfrak{U} \setminus (\mathfrak{U} \times \mathfrak{U}))$ would be a set using the fact that $\mathfrak{O}$ is a definable proper class, and applying selection.)  On the other hand, $\mathfrak{O}$ cannot be a set because of the Burali-Forti paradox, giving a contradiction.
