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Let $\kappa$ a cardinal of cofinality $\omega$; let $C \subseteq \kappa$ be a unbounded countable subset. Why is then $C$ closed (and thus a c.u.b.)? This means that if $\delta < \kappa$ is a limit ordinal, and $C \cap \delta$ is unbounded in $\delta$, then $\delta \in C$.

I doubt that this is true; however, it's a short statement in Kunen's set theory book.

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What Kunen meant was that if $C$ is the range of a countable increasing unbounded sequence in $\kappa$, that is, $C$ is unbounded and has order type $\omega$, then it will be vacuously closed, since it has no limit points below $\kappa$.

This situation shows that the concept of club on cardinals of countable cofinality doesn't work as we want.

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    [Do you think I should post further questions like that on mathoverflow, or keep them here?]2010-09-02
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    Martin, I've enjoyed your MO posts over there, so it seems that you have a good idea on which forum is appropriate for which post.2010-09-02
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If you set $\kappa = \aleph_{\omega}$ and let $C = \{0,,1,\aleph_1, 2, \aleph_2,...\}$ (so $C$ is the union of all natural numbers together with all $\aleph_n$ for $n>0$) and you let $\delta = \omega$, I think you have a counterexample.

$C\cap\delta = \delta$, so is automatically unbounded in $\delta$. But $\delta$ itself isn't in $C$ by construction.

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    Thank you. It wonders me that Kunen claims this, but on the same page there is your example (showing that club sets don't yield a filter).2010-09-02