I'm new to set theory and can't find my way through this. I've defined $\omega_1$ as the Hartogs number of $\omega$, that is $\omega_1:=H(\omega)=\{\alpha\in On: |a|\leq\omega\}$. I believe contradiction is the only way through this, so let's suppose that $\omega_1=\beta+1$ for some ordinal $\beta$. Then $\beta<\omega_1$, therefore $\beta\in\omega_1,$ and consequently $|\beta|\leq\omega$. But then $\beta+1$ is at most countable, a contradiction, since $\omega_1$ is the first non-countable ordinal.
Now my question has two parts:
1) Is the proof above correct?
2) Is it true that $\omega_{a+1}:=H(\omega_a)$ is a limit ordinal for all $a\in On$? If so, can you provide a proof description since the above cannot be modified (in the initial situation we had the benefit of $\omega$ being countable)
Note that for an infinite ordinal, $|\alpha|=|\alpha+1|$. The proof is literally the same as proving that adding $-1$ to the natural numbers results in a countable set.
This means that if $\alpha$ is countable, so is its successor. In particular, the least uncountable ordinal cannot be a successor ordinal.