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.

  • $\begingroup$ Hint on #2: it has a much cleaner characterization than the written one...(Secondary hint: what is $\omega+\epsilon_0$?) $\endgroup$ May 17, 2016 at 16:56
  • $\begingroup$ I'm not sure what $\epsilon_0 $ is $\endgroup$
    – Jytug
    May 17, 2016 at 16:57
  • $\begingroup$ In that case, $\omega + \omega^\omega$? $\endgroup$ May 17, 2016 at 16:58
  • $\begingroup$ So by $ \epsilon_0 $ you mean the first ordinal such that $ \omega + \epsilon_0 = \omega $? And for sufficiently large $ \alpha $ it holds that $ \omega + \alpha = \alpha $? $\endgroup$
    – Jytug
    May 17, 2016 at 17:12
  • $\begingroup$ $\epsilon_0$ has a different definition than that; it was just a random 'large countable ordinal' pulled out of a hat. But your statement that $\omega+\alpha=\alpha$ for large ordinals is basically correct; more broadly, can you characterize all the ordinals that are of the form $\omega+\alpha$? Cantor normal form will be your friend here... $\endgroup$ May 17, 2016 at 17:18

1 Answer 1


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.

  • $\begingroup$ But 1. contains $\{\alpha \in \omega_1 \mid Lim(\alpha)\}$, which is club in $\omega_1$. $\endgroup$
    – BrianO
    May 17, 2016 at 20:52
  • $\begingroup$ But it's not a club in $\omega_2$, because $\omega_1<\omega_2$ and therefore it is bounded there. $\endgroup$
    – Asaf Karagila
    May 17, 2016 at 21:00
  • $\begingroup$ Sure, it isn't club in $\omega_2$, but I thought th equestion might be a bit of a "trick" question in that it doesn't specify "club in what?". $\endgroup$
    – BrianO
    May 17, 2016 at 21:20
  • $\begingroup$ Oh, c'mon, now you're being intentionally... a mathematician. :-P $\endgroup$
    – Asaf Karagila
    May 17, 2016 at 21:23
  • $\begingroup$ Haha! Probably. Or pointlessly cautious, or paranoid ;) $\endgroup$
    – BrianO
    May 17, 2016 at 21:50

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