# 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?

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