How to prove that a collection of elements is a class which is not a set. I am reading a set theory course, which deals with ordinals, classes, transfinite recursion.
As an exercise of recursion, it is given that: $\mathord{\in} = \{ \langle x,y \rangle ;x \in y\}$. It is asked to prove that $\mathord{\in} \subseteq V \times V \setminus \{ \varnothing \}$. It is also asked to show that, $\mathord{\in}$ is a class which is not a set.
I have a few problems: first, I am not sure what is the connection to recursion. Second, I am not sure, how can one prove that a collection is a class? Also, isn't it obvious that $\mathord{\in} = \{ \langle x,y \rangle ; x \in y\}$?
Thank you,
Shir 
 A: *

*The connection to recursion is presumably the observation that one can define functions by transfinite recursion along well-orderings that are proper classes of ordered pairs. However, a point that does not come up in your question is that to do this, every proper initial segment of the well-ordering must be a set.

*Here is another proof that the class $R_\in = \{(x,y): x\in y\}$ is not a set.    If $R_\in$ were a set, then its image under the class function $(x,y) \mapsto x$, that is, under projection in the first coordinate, would be a set by Replacement.  However, as Asaf points out, the image of $R_\in$ under this projection function is $V$ because every set $x$ is an element of another set $\{x\}$.  Then one uses the fact that $V$ is a proper class to get a contradiction. This proof is independent of our particular definition of ordered pairs, and I think it is the proof that generalizes best: To figure out whether a class is a set or a proper class, use injections or surjections to compare its size to that of a known set or known proper class and then use Replacement.

*It is not literally true that $\mathord{\in} = \{(x,y): x\in y\}$.  Rather, this is an abuse of notation.  The "$\in$" symbol denotes a relation symbol in the language of set theory, not a variable symbol.  This abuse of notation makes sense by analogy with writing $R = \{(x,y): x \mathbin{R} y\}$ where $R$ is a variable symbol that denotes a relation, which is itself an abuse of notation for the opposite reason.  One bad thing about using "$\in$" as a variable symbol is that one has to write \mathord{\in} in order to get the spacing right in $\LaTeX$.  This is why I used $R_\in$ above instead.
A: Classes are definable collections (in the language of set theory), so just by writing it you prove that it is a class. If ordered pairs bother you, then you can write instead any way of coding ordered pairs with sets.
To show that this is not a set you can show that every set appears in the left coordinate of an ordered pair of $\in$. Therefore the function mapping an ordered pair to its left coordinate is a surjection. If you use the Kuratowski definition for ordered pairs then $\langle x,y\rangle = \{\{x\},\{x,y\}\}$ and so $\bigcup\bigcup\in=V$.
A: Another way to show that it is not a set is to note that if it were a set, then we could use it as a basis for forming other sets, in particular the ordered pair $\langle \mathord{\in} , \{ \mathord{\in} \} \rangle$.  Note now that $\langle \mathord{\in} , \{ \mathord{\in} \} \rangle$ would be an element of $\mathord{\in}$.  But then $$\mathord{\in} \in \{ \mathord{\in} \} \in \{ \{ \mathord{\in} \} , \{ \mathord{\in} , \{ \mathord{\in} \} \} \} \in \mathord{\in}.$$  I think there's a contradiction1 somewhere.

1Assuming Foundation
