Is {x : ϕ(x)} a set with ϕ(x) being the property: ∀y (x ∈ y)

Question

The property ∀y (x ∈ y) contradicts the axiom of regularity since x ∈ x is not true for any non-empty set x. Does this mean that {x : ϕ(x)} = $\emptyset$ which is a set or because it violates an axiom does that mean it is not a set?

Answer 1

In ZF $\mathbb\{x\mid \forall y(x\in y)\}$ is a set and - as you noticed - it is the empty set.

Actually we can prove for some set $a$ that $\mathbb\{x\mid \forall y(x\in y)\}=\mathbb\{x\in a\mid \forall y(x\in y)\}$ and then the axiom scheme of separation ensures us that we are dealing with a set.

Answer 2

The former.   The predicate is false because of the axiom, but it does not violate any axiom to build a set with a false constructor.   You just build the emptyset.