Is a class of sets a set if its union is a set? I am wondering if a class of sets X is a set given that $ \cup X $ is a set.
Going in the other direction is trivial -- if X is a set then $\cup X$ is a set by the axiom of union. However, how would one prove it in this direction?
My initial thought is that if $\cup X$ is a set, we can construct its power set $P(\cup X)$. Then, to show X itself (the class of sets) is a set, it would suffice if it were a member of $P(\cup X)$. I'm not sure if this is true or if it is, how I would show that it is?
 A: $X$ is not necessarily a member of $P(\bigcup X)$ but rather is a subclass of it.  Indeed, if $x\in X$, then $x$ is a subset of $\bigcup X$ (since $\bigcup X$ is the union of all the elements of $X$), so $x\in P(\bigcup X)$.  So $X$ is a subclass of $P(\bigcup X)$ and hence is a set by Separation.
A: The real question is how you can define the class of sets in the first place without knowing that it is a set. Presumably you have some formula $\varphi$ that defines the members of the class, and you can prove
$$\exists x\forall y\big(y\in x\leftrightarrow\exists z(\varphi(z)\land y\in z)\big)\,;$$
that is, you can prove that there is a set that behaves like the union of the class of sets satisfying $\varphi$. In that case you can let $a$ be such a set $x$, so that
$$\forall y\big(y\in a\leftrightarrow\exists z(\varphi(z)\land y\in z)\big)\,,\tag{1}$$
and define a set
$$X=\{y\in\wp(a):\varphi(a)\}\,;$$
Clearly every member of $X$ satisfies $\varphi$. Now suppose that $z$ is a set, and $\varphi(z)$ holds. Then $(1)$ implies that $\forall y\in z\,(y\in a)$, i.e., that $z\in\wp(a)$, and hence that $z\in X$, so $X$ is indeed the set of sets satisfying $\varphi$.
