Can a definite condition be considered as an object in $ZF$?

Question

Can a definite condition be considered as an object in $ZF$?

This question arose from the following question:

For every class $A$, prove that $A$ is a set if for some class $B, A \in B$

Where $A \in B$ is defined to be equivalent to the fact that $B$ is a set and $A$ is a member of $B$, or (if $B$ is a unary definite condition) $B(A)$.

If anyone can answer both that'd be great

Answer 1

Well. Not quite exactly, but almost.

For example, the empty set is really the class $\{x\mid x\neq x\}$. But really there is an axiom which states $\exists x(\forall y(y\in x\leftrightarrow y\neq y))$.

So a condition which defines a class is not a set per se, but it can be extensionally equivalent to a set.

For another example, if $x$ is a set, then the class $\{y\mid y\in x\}$ is equivalent to the set $x$.