Show that $\omega + \omega$ is countable.

Question

Two sets are equinumerous if there exists a bijection between them. A set is finite if it is equinumerous to some $n \in \omega$ and a set is countable if it is equinumerous to $\omega$.

$\omega$ is the set of all natural numbers as an ordinal.

How can I prove $\omega+\omega$ is countable? Where $+$ is the ordinal addition.

Answer 1

Hint: $$ \omega+\omega=\omega\cup\{\omega+n:n\in\omega\} $$ and $\omega=\{2n:n\in\omega\}\cup\{1+2n:n\in\omega\}$

Answer 2

ω + ω is like the set of all ω + n for any n ∈ ℕ. So to map it onto ω we would map ω onto 0, ω + 1 onto 1 and so on, seeing as there’s a bijection, it must be countable.