# 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?

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.