Is it consistent with $\sf ZF-Fnd$ that no class whose union is the whole universe can be well-ordered?

If $R$ is a binary relation; $\phi $ is a unary predicate; then:

$ \neg [R \text { well orders } \phi \land \forall y \, \exists x: \phi(x) \land y \in x ]$

Where: $R \text { well orders } \phi \equiv_{df} \\\big{(}\forall x \,\forall y \, \forall z: \phi(x) \land \phi(y) \land \phi(z) \to \\ (x R y \to \neg y R x) \land \\ (x \neq y \to x R y \lor y R x) \land \\ (x R y R z \to x R z) \land \\ \exists u \forall m (m \in u \iff \phi(m) \land m R x)\big{)} \\\land \\\forall s [(\exists r \in s : \phi(r)) \to \exists k \in s( \phi(k) \land \forall v \in s : \phi(v) \to \neg v R k)] $

In English: No class whose union is the universe can be well orderable?

Now this is an anti-foundation axiom, since we can take $\phi$ to be a stage of the cumulative hierarchy which is clearly well orderable.

Is that principle consistent with ZF-Fnd?

Solution 1:

Sure.

Suppose that there is a proper class of atoms, but every set of atoms is finite. Easily, there is no well-ordering of the whole universe, since that would imply we can pick some infinite set of atoms.

On the other hand, if some class whose union is the whole universe was well-orderable, note that every set must contain at most finitely many atoms. So recursively we can go about picking the first $\omega$ sets which contain new atoms, and this will define, again, an infinite set of atoms.