I am trying to prove the following:
- $cf(\omega)=\omega$.
- $cf(\alpha)<\omega$ iff $\alpha$ is successor ordinal.
For 1. define a cofinal map follows $f\colon \omega\to\omega$ by $f(n)=n+1$. It is clearly unbounded. To argue it is the smallest. Let $m<\omega$ and consider cofinal map from $m\to\omega$ but in this case we will have $$\omega=\bigcup\{f(n)+1\colon n\in m\}$$ It means $\omega$ is finite which is a contradiction since $\omega$ is countable ordinal.
For 2. One direction is obvious. So, assume $cf(\alpha)<\omega$ and we want to show $\alpha$ is successor. $cf(\alpha)=m$ for some $m\in\omega$. So there exists a cofinal map $f\colon m\to\alpha$. As in the previous proof $$\alpha=\bigcup\{f(n)+1\colon n\in m\}$$ So finite union of successor ordinals is successor ordinals (NOT sure).
Please tell me if what I did is correct or not. Also,