# Does every countable subset of the set of all countable limit ordinals have the least upper bound in it?

I'm sorry if the question is that kind of trivial, I just feel uncertain about these ordinals all the time. Is the answer to the following question "yes":

Denote by A the set of all countable limit ordinals. Does every countable subset of A have the least uper bound in A?

## 1 Answer

Assuming the axiom of choice, the countable union of countable sets is countable. Therefore the countable limit of countable ordinals is also countable. And trivially, the limit of limit ordinals is a limit ordinal.

So the answer is yes. The set of countable limit ordinals is closed under countable sequences.

However if we don't assume the axiom of choice, then it is consistent that $\omega_1$ is the countable limit of countable ordinals, in which case the answer would be negative.

• Oh, mmm, well, I'm sorry for the question perhaps being a kind of ambiguous, let me clarify: denote by A the set of all countable limit ordinals. Does then any countable subset of A have the least upper bound in A (I mean, not in itself... since $\omega^2$ is still countable, and limit)? – W_D Jul 7 '13 at 11:38
• @Alex: Please edit that into the question, so other people can read as well. – Asaf Karagila Jul 7 '13 at 11:39