Non-aleph infinite cardinals

Solution 1:

The class of $\aleph$ numbers is the same class of cardinals that you know in a model of ZFC. Namely, initial ordinals. The definitions are exactly the same. Furthermore by definition the $\aleph$ cardinals are ordinals, so the correspond to well ordered sets.

On the other hand, if $A$ is not a well-orderable set, then $|A|$ corresponds to the set $$\{B\mid \exists f\colon A\to B\text{ a bijection}\land\operatorname{rank}(B)\text{ is minimal}\}$$ Where the $\operatorname{rank}$ operator is the von Neumann rank of $B$. This set is not an ordinal, clearly, and it may lack any internal structure.

The class of cardinals, therefore, is combined from two parts:

1. The $\aleph$ numbers which are "ordinal which cannot be put in bijection with any of its elements".

   We can see that the $\aleph$ numbers do not form a set directly, suppose that they would, then there was an ordinal $\gamma$ such that the set of $\aleph$ has von Neumann rank $\gamma$. In particular all of its elements have rank $<\gamma$. Let $\kappa$ be the first ordinal above $\gamma$ such that $\kappa$ is not in bijection with any of its elements, then $\kappa$ is an $\aleph$, but its von Neumann rank is $\kappa>\gamma$ in contradiction.

2. Cardinals of sets which are not well-orderable. These are described as sets $A$ such that "Every two members of $A$ have a bijection between them, all the elements of $A$ have the same von Neumann rank, and no set of lower rank has a bijection with any element of $A$, and if there is a $B$ of the same von Neumann rank as a member of $A$, and they are in bijection then $B$ is an element of $A$ as well"

   Yes, it is a bit clumsy and unclear, but set theory without choice may get like that often.

It is immediate that the class of cardinals is a proper class since it contains all the $\aleph$-cardinals. Much like in ZFC the cardinals make a proper class, the arguments carry over in this case as well.

Lastly, you cannot prove that a power set of a well-ordered set is well-ordered because if the axiom of choice fails this is simply not true. Furthermore, $A$ itself is a class, as it contains elements of unbounded rank, so we need to be more careful with "the union over $A$" as it is not a set as well, that is $V$ itself is a class.

As $V$ is a class its power "set" is not a set and does not exist, and as I remarked power sets of a well-ordered set need not be well-orderable.

See also:

1. Defining cardinality in the absence of choice
2. There's non-Aleph transfinite cardinals without the axiom of choice?
3. How do we know an $ \aleph_1 $ exists at all? (this asserts that $\aleph_1$ exist, even without choice, and the argument carries over to high cardinals)

Solution 2:

The well-ordered cardinals are defined by the formula $x\in\text{ON}\land |x|=x$. Of course both conjuncts abbreviate fairly complicated formulae, but they’re familiar ones.

As for the first question, the $\omega_\alpha$ are defined by transfinite recursion and are independent of AC.