$|\omega+\omega|=|\omega\cdot\omega|=|\omega^\omega|=|\omega|$
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
We have:
$\omega+\omega=\{\omega+\alpha\mid \alpha<\omega\}$
$\omega\cdot\omega=\{\omega\cdot\alpha \mid \alpha<\omega\}$
$\omega^\omega=\{\omega^\alpha \mid \alpha<\omega\}$
Thus we build bijections $f,g,h$ as follows:
$f:\omega \to \omega+\omega$ such that $f(\alpha)=\omega+\alpha$
$g:\omega \to \omega\cdot\omega$ such that $f(\alpha)=\omega\cdot\alpha$
$h:\omega \to \omega^\omega$ such that $f(\alpha)=\omega^\alpha$
This completes the proof.
Since $\omega$ × $\omega$ equinumerous to N×N, it is countable.
By induction for all n in N, $\omega^n$ is countable.
$\omega^{\omega}$ = sup$_n$ $\omega^n$ = $\cup_n \omega^n$ is a countable union of countable sets equinumerous to N×N, hence countable.
Finally as $\omega$ + $\omega$ < $\omega$ × $\omega,$ it too is countable.