Can all ordinals be obtained by taking successors and countable limits? Define a class $K$ of ordinals inductively as follows:


*

*$0=\emptyset\in K$.

*For all $\alpha\in K$, the succesor of $\alpha$ is also an element of $K$.

*For every function $f\colon \mathbb N\to K$, the ordinal that immediately follows after all ordinals $f(0),f(1),\dots f(n),\dots$ is also an element of $K$.
Call an arbitrary ordinal $\beta$ fett iff $\beta\not\in K$.
Question. Do fett ordinals exist? In other words:

Is there an ordinal $\beta$ such that $\beta$ is not an element of the class $K$?

 A: First of all, note that the class of all the ordinals satisfies this definition. You want to say that $K$ is the smallest class of ordinals satisfying that the three requirements hold.
And I claim that at least under the axiom of choice $\omega_1$ is such a class of ordinals, and therefore $\omega_1$ is a Fett ordinal. It is in fact, the least Fett ordinal, therefore making it the Boba Fett of ordinals.
Let's see why.


*

*$\varnothing\in\omega_1$, easy peasy.

*If $\alpha\in\omega_1$, then $\alpha+1$ is also countable, so $\alpha+1\in\omega_1$ as well.

*Finally, if $f\colon\Bbb N\to\omega_1$ is any function, then $\sup\operatorname{rng}(f)$ is the countable union of countable ordinals, therefore a countable ordinal. So we have all three requirements.


However, the axiom of choice is needed here. Not only that it is consistent that $\omega_1$ is not not a Fett ordinal without the axiom of choice, namely it can be the countable union of countable ordinals; it is possible that every ordinal is a successor or a countable union of smaller ordinals. In that case, no ordinal is a Fett ordinal.
So you need the axiom of choice to conclude the existence of Boba Fett.
A: Yes, such ordinals exist - for example, $\omega_1$, the first uncountable ordinal. 
The crucial issue here is cofinality: the cofinality of an ordinal $\alpha$ is the least $\beta$ such that there is a function $f:\beta\rightarrow\alpha$ whose range is unbounded in $\alpha$. It's easy to show that $K$ consists exactly of those ordinals of countable cofinality - that is, whose cofinality is $\omega$ - which are also not larger than any ordinal of uncountable cofinality.
Now, $\omega_1$ is not of countable cofinality, since the union of countably many countable sets is countable. So $\omega_1$ - and any larger ordinal - is not in $K$.
