Proof its proper class I am reading set theory course, and there were 2 tasks to prove $T_1 = \{ x |\emptyset\in x \}$ and $T_2 = \{x| \emptyset \notin x\} $ are proper classes using Axiom schema of replacement and Axiom of union.
The task is really confusing for me, because how can both classes be proper classes if both statements are the opposites?
Could you please help me?
 A: Intuitively, a set is a proper class if it is "too big." A good first step towards developing an intuition for proper classes (although one you don't want to push too far) is to consider the case of finite vs. infinite sets in the natural numbers. Certainly it's possible for both a set of natural numbers and its complement in the natural numbers to be infinite: consider e.g. the set of even numbers and the set of odd numbers. The takeaway from this example is:

Just because the classes $T_1:=\{x: \emptyset\in x\}$ and $T_2:=\{x: \emptyset\not\in x\}$ are complementary doesn't mean that they can't both be "too big."

In fact, a good step towards understanding why they're both big is to convince yourself that they have the same size. Specifically, there is a simple bijection$^*$ between them:

 Given a set $x$ in $T_2$, think about $x\cup\{\emptyset\}$; or, in the other direction, given a set $y\in T_1$ think about $y\setminus\{\emptyset\}$.

$^*$OK fine, this isn't properly speaking a bijection because it's not even a function in the first place: functions, after all, have to be sets and what we're trying to prove is exactly that $T_1$ and $T_2$ are "too big" to be sets. Really it's a "class bijection." But since the above is really just for developing intuition, this point is worth ignoring for now.

The above should convince you that either $T_1$ and $T_2$ are each sets or $T_1$ and $T_2$ are each proper classes. So now the question is: which option is it?
There are a few ways to approach this, but I think the simplest is the following. The union of two sets is always a set again (this is a good exercise if you haven't already proved it); now, given that $T_1$ is exactly the "opposite" of $T_2$, what can you say about $T_1\cup T_2$? Is the result something you already know is not a set?
This sequence of ideas - "$T_1$ and $T_2$ have the same size and their union is a proper class, so they can't both be sets, which means they must both be proper classes" - is an excellent blueprint for an actual proof. It's not quite a proof on its own since it's a little too vague, but it points the way to a perfectly correct argument. As a hint, try to set things up as a proof by contradiction; assuming $T_1$ and $T_2$ are sets, you'll be in a good position to use Replacement and Union.
