I'm getting acquainted with the stationary sets and clubs. I don't yet quite get everything well, so I'd appreciate some help with this question:
which of the following sets are clubs, contain a club, are stationary?
1. $ \{ \alpha \in \omega_2 : \alpha \in \text{Lim} \text{ and } \text{cf}(\alpha) = \omega \}$
2. $ \{\alpha \in \omega_2: \exists~ \beta \text{ s.t. }\alpha = \omega + \beta \}$
3. $ \{ \alpha \in \omega_1 : \text{ot}(\alpha \cap \text{Lim}) = \alpha \}$
Where $ \text{cf, ot, Lim} $ denote, respectively, cofinality, order type and the class of limit ordinals.
Here are some of my thoughts:
1. I believe that $ \sup\limits_{\omega \leq \alpha < \omega_1} \alpha = \omega_1$ and $ \text{cf}(\alpha) = \omega $, which would mean that this set is not closed. I'm not sure, however, how to decide if it contains a club. I know that it's stationary, since for any club $ C $ we can take a strictly increasing countable sequence $ \{c_n\} \subset C $ and its limit belongs to the set.
Is this correct? As for 2 and 3 I am not really sure where to begin. I would appreciate some hints.
Your argument about (1) is correct, and it is really the reason that it cannot contain a club. If $\kappa>\omega$ is regular, and $D\subseteq\kappa$ is a club, then it must have limit points of every cofinality below $\kappa$ (alternatively, simply note that $\{\alpha<\omega_2\mid\operatorname{cf}(\alpha)=\omega_1\}$ is stationary, and therefore the set in (1) cannot be a club.)
In the second case, try to find out which ordinals---in general---cannot be written as $\omega+\beta$ for some $\beta$. You'll find out that there are only countably many of them.
For the third one, it is not hard to prove that the set is closed; to see it is unbounded look at $\varepsilon_0$ and other $\varepsilon$-numbers, and try to understand whether or not they belong to this set.