Ackermann set theory appears to prove inaccessible cardinals exist?

Question

I know this proof must be wrong, as it would mean that ZF proves inaccessible cardinals exist, which it doesn't.

Let A(x) = ∀I, I is an inaccessible class ⇒ x∈I

Now the class of all sets V must be an inaccessible class, if it was not then it would be a set, but that is a contradiction.

So A(x) ⇒ x∈V, so by the Axiom of set comprehension { x ∈ V | A(x) } is an inaccessible set.

I think it's either that limitation of size doesn't hold, so V can be the union of a class of less than V sets, while still being a proper class, or the intersection of every inaccessible class is not an inaccessible class.

Edit: I've realised limitation of size can't be true in Ackermann set theory, so I think it's just possible for V to be singular.

If it's the case that V can be singular, could this limitation of size style axiom make the proof go though.

Lim: If X is a class containing only sets, and there is no injection V to X, then X is a set.

Answer 1

Ackermann set theory doesn't prove that $V$ is inaccessible. So $V$ can indeed be singular. Anyhow if you add the axiom of limitation of size on $V$, i.e. every subclass of $V$ that is strictly smaller than $V$ is an element of $V$, then of course the resulting theory would prove $V$ to be inaccessible, and would be STRONGER than $ZFC$.

Answer 2

For a family of sets $\mathcal F$, if you want to construct their intersection $\bigcap \mathcal F$, you must first show that $\mathcal F$ is non-empty.