Forgive me if this is a silly question. I am a little stuck on a definition from Kunen's text.
He says:
Definition 10.3. If $A$ can be well-ordered, $|A|$ is the least $\alpha$ in bijection with $A$.
My confusion is as follows: If we assume choice, then every set is well-orderable. i.e. for all $x$ there exists $\alpha$ in bijection with $x$, such that the bijection is moreover an order isomorphism. However, it seems to me then, that choice acts as a class function on $\mathbf{V}$ assigning to each set an order-type from those ordinals in bijection with it, guaranteeing also that there is at least one such ordinal. Otherwise, I think Kunen's definition is pointless, for otherwise we may always select the order-type of $x$ to be the least ordinal in bijection with it and then just set the cardinality to be the order type naturally induced by the aforementioned bijection.
Is this a correct way of viewing choice? Or am I muddling something up?
If $A$ is well-ordered, then we have a unique ordinal which is its order type, and in fact there is a unique bijection which is an order isomorphism. Otherwise, it can be, but then it may have many different order types (in fact, if $A$ is infinite, it is exactly what is going to happen).
Moreover, even if you fix one bijection between $A$ and some ordinal, we can apply permutations of $A$ or permutations of the ordinal to get different bijections. So there are many different bijections possible. Which one do we use? It isn't clear.
To see a similar situation, note that there is no canonical way to choose how to enumerate each countable subset of $\Bbb R$, unless of course we well-ordered $\Bbb R$ and choose an enumeration for each countable ordinal, and then used that to argue that each countable set is isomorphic to a unique countable ordinal (as a subset of our fixed well-order), and now appeal to the bijection we chose between that ordinal and $\omega$.
And indeed, without the axiom of choice, it may very well be that we just cannot make these choices, and even though every countable subset of $\Bbb R$ is well-orderable (indeed, it is countable!), we still cannot assign them all, at the same time, bijections with $\omega$.
The same holds in the case of proper classes and "just $\sf ZFC$". In other words, it is quite possible for a model of $\sf ZFC$ to exist in which there is no definable way to assign a well-ordering to each set. This is exactly the situation when the Axiom of Global Choice fails.