Subclass of ordinals is a set iff it is bounded Let $X\subseteq \text{Ord}$. Then $X$ is a set if and only if there exists an ordinal $\beta$ such that for all $\alpha \in X$, $\beta\ge \alpha$.
I am really having trouble with proving that something is a set or using the fact that we have a set. How does one approach problems like this, when you have to prove something is a set?
 A: It makes sense to take the axiom system you're working with (ZF I suppose?) and go from there. "x is a set" is just short for "ZF proves the existence of x". Depending on your formulation of the ZF axioms, one would approach this as follows: for example, to show that for sets $x,y$ the Kuratowski pair $\langle x,y \rangle = \{ \{ x \} , \{ x,y \} \}$ is a set, we apply pairing twice to get a set $z$ that contains $\langle x,y \rangle$, and then use comprehension on $z$ to obtain the Kuratowski pair itself. Note we have assumed the existence of arbitrary sets and only used the axioms from there to derive the conclusion.
In order to show that some collection is in fact a proper class and hence not a set, a proof by contradiction is usually the way to go. In your particular example, indeed proceed by contradiction. If $C$ is an unbounded class of ordinals and we assume it is a set, is there an axiom of ZF that proves the existence of a set containing each ordinal in $C$? What is this set equal to?
