# An upper bound of a countable set of countable ordinals

Let $X$ be a countable set of countable ordinals. Is there any upper bound of this set under $\omega_1$? In other words, can we find an ordinal $\alpha<\omega_1$ which is larger than any elements of $X$?

HINT: If $X$ is a set of ordinals, then $\bigcup X=\sup X$. If $X$ is a countable set of countable ordinals, then what is the cardinality of $\bigcup X$?
• Fairly certain you don't need Choice here. The countable union of countable sets isn't countable in general without Choice, but that's because you need to order each of the sets (put them in 1-to-1 correspondence with $\omega$). Here, you can use the fact that they're well-ordered to define a good ordering, and you don't need Choice... I thought. – Akiva Weinberger Mar 9 '15 at 9:43
• Yes, yes you need the axiom of choice here. $\omega_1$ is consistently singular without the axiom of choice. Meaning that it is the countable union of countable ordinals. Classical examples are the Feferman-Levy model and the Truss models for the perfect set property from a singular cardinal. Other models include Prikry forcing over models where $\omega_1$ is measurable (e.g. models of $\sf ZF+DC+AD$). – Asaf Karagila Mar 9 '15 at 9:44
• You mean $\bigcup X$ is countable and it is contained in another countable ordinal, right? I have thought about that, but I have no idea about showing the existence of such countable ordinal containing $\bigcup X$.. – Therefore.. Mar 9 '15 at 13:41
• @Ifexists: No, $\bigcup X$ is an ordinal, and assuming the axiom of choice, it's also countable. Just show that $\bigcup X$ satisfies the definition of an ordinal (whichever definition you have), and show that if $\beta$ is an upper bound for $X$, then $\bigcup X\leq\beta$ to conclude that this is the supremum. – Asaf Karagila Mar 9 '15 at 13:52