Prove the limit case: if $\alpha,\beta$ are countable ordinals, then $\alpha+\beta$ is also countable

Question

I am trying to prove that if $\alpha,\beta$ are countable ordinals, then $\alpha+\beta$ is also countable.

Both base and successor case has been done, now I am considering the limit case. Once I assume that $\forall \beta< \gamma, \alpha+\beta$ is countable, what should I do to conclude that $\sup\lbrace \alpha+\beta\mid\beta<\gamma\rbrace$ is also countable?

Thanks in advance!

EDIT: The version of definition of ordinal I use here: An ordinal is a set $\alpha$ such that:

Answer 1

You can do it by taking the synthetic definition of addition. Let $A$ be a set well-ordered by an ordering of length $\alpha$, and $B$ likewise $\beta$. Then the ordinal sum $\alpha + \beta$ is given by ordering the set $A \sqcup B$ (where $\sqcup$ denotes the disjoint union) in a certain way so as to place a copy of $B$ after a copy of $A$. That set is countable, being the union of two countable sets.