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Let $\kappa$ be a singular cardinal with $\operatorname{cf}\kappa = \lambda > \omega$.

Let $C = \{ \alpha_{\zeta} \mid \zeta < \lambda \}$ be a strictly increasing continuous sequence of cardinals with limit $\kappa$. I want to show this is a closed unbounded set in $\kappa$.

Obviously it is unbounded, because $\sup C = \kappa$, but how do I show it is closed? Is it to do with continuity? Also I am used to dealing with clubs in regular cardinals, are there any differences for clubs in singular cardinals?

Thanks very much.

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2 Answers

up vote 4 down vote accepted

Yes, you are correct. The fact that $C$ is closed has everything to do with the sequence $\{ \alpha_\zeta \}_{\zeta < \lambda}$ being continuous.

The main idea is that since the sequence is strictly increasing, then for any limit ordinal $\gamma < \lambda$ we have that $\lim_{\zeta < \gamma} \alpha_\zeta = \sup ( \{ \alpha_\zeta : \zeta < \gamma \} )$. Therefore, if $\alpha < \kappa$ is a limit ordinal such that $\sup ( C \cap \alpha ) = \alpha$, let $\gamma = \min \{ \zeta < \lambda : \alpha_\zeta \geq \alpha \}$. We can show that $\gamma$ must be a limit ordinal, and that $C \cap \alpha = \{ \alpha_\zeta : \zeta < \gamma \}$. Using continuity we have that $$\alpha_\gamma = \lim_{\zeta < \gamma} \alpha_\zeta = \sup ( \{ \alpha_\zeta : \zeta < \gamma \} ) = \sup ( C \cap \alpha ) = \alpha,$$ and so $\alpha \in C$.

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Thanks, but why is $\gamma$ a limit ordinal? I can't find a contradiction if I suppose it is a successor. Even it if it's not a limit ordinal, surely $\alpha_{\gamma} = \lim_{\zeta \to \gamma} \alpha_{\zeta}$ still holds? –  Paul Slevin Feb 27 '12 at 15:12
No sorry, that is wrong. If $\gamma = \beta + 1$, then $\alpha_\beta = \lim_{\zeta \to \gamma}\alpha_{\zeta}$. I think the answer to my question follows after showing that $C \cap \alpha = \{ \alpha_{\zeta} \mid \zeta < \gamma \}$, and then we get a contradiction if we assume $\gamma$ is a successor. –  Paul Slevin Feb 27 '12 at 15:22
@Paul: You're correct that the contradiction will follow once you show that $C \cap \alpha = \{ \alpha_\zeta : \zeta < \gamma \}$. This little observation stems from the definition of $\gamma$ and the fact that the original sequence is strictly increasing. –  Arthur Fischer Feb 27 '12 at 16:24
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Recall the definition of a closed and unbounded set in $\kappa$:

  1. For every $\alpha<\kappa$ there is some $\alpha_\zeta\in C$ such that $\alpha<\alpha_\zeta$ (unboundedness);
  2. For every limit ordinal $\delta<\lambda$ we have $\sup\{\alpha_\zeta\mid\zeta<\delta\}=\alpha_\delta$ (continuity).

Now if $\sup C=\kappa$ then it is clearly unbounded in $\kappa$, and continuity is exactly the demand that the sequence is closed. As you may recall, a continuous sequence is the range of a continuous function with respect to the order topology.

As for clubs of singular cardinals, those are indeed a bit different. In regular cardinals being a club means being a very large set in every possible way, while clubs in singular cardinals can be quite small in terms of cardinality.

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I forget, Paul, did we discuss the continuity in the previous answer or on the chat? –  Asaf Karagila Feb 24 '12 at 16:35
you defined a new sequence in your previous answer, which was continuous –  Paul Slevin Feb 27 '12 at 14:12
Or if you mean in terms of continuity in continuous functions, we talked about it in chat a little bit –  Paul Slevin Feb 27 '12 at 14:28
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