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I'm studying the ordinal space $[0,\kappa[$ where $\kappa\neq \omega$ is a cardinal of countable cofinality and I want to know why there are in $[0,\kappa[$ two disjoint closed sets of cardinality $\kappa$.

The case when $\kappa=\omega$ it's obvious but I can't prove the general one... If anybody could help me I would be more than gratefull.

Thanks!

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  • $\begingroup$ $\kappa=\omega+1$ is not a cardinal number... $\endgroup$ – Ergonvi Jan 5 '15 at 10:04
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    $\begingroup$ Didn't you ask that on math.SE before? $\endgroup$ – Asaf Karagila Jan 5 '15 at 10:04
  • $\begingroup$ @Ergonvi -- sorry, I removed my wrong comment before I saw your correction (you were fast! :-). I overlooked the phrase countable cofinality. $\endgroup$ – Włodzimierz Holsztyński Jan 5 '15 at 10:07
  • $\begingroup$ @Ergonvi -- and yes, I overlooked wordcardinality too. I had overlooked everything! $\endgroup$ – Włodzimierz Holsztyński Jan 5 '15 at 10:12
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Let $\kappa_n$ be increasing cofinal in $\kappa,$ and let

$X=\bigcup_{n}(\kappa_{2n}, \kappa_{2n+1}]$, and $Y=\bigcup_n (\kappa_{2n+1}, \kappa_{2n+2}]$

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  • $\begingroup$ Thank you @Mohammad Golshani ! But why $X$ and $Y$ have cardinality $\kappa$? $\endgroup$ – Ergonvi Jan 5 '15 at 10:03
  • $\begingroup$ That's trivial, for example $|X|=\Sigma_n \kappa_{2n+1}=\kappa.$ Similarly $|Y|=\Sigma_n \kappa_{2n+2}=\kappa.$ $\endgroup$ – Mohammad Golshani Jan 5 '15 at 10:25
  • $\begingroup$ Oh, that's right... Thank you @Mohammad Golshani! :) $\endgroup$ – Ergonvi Jan 5 '15 at 17:59
  • $\begingroup$ But the same argument wouldn't be right for cardinals with uncontable cofinality? I mean for example, $X=\bigcup_{\alpha <\omega_1} (\kappa_{2\alpha},\kappa_{2\alpha+1}]$ and $Y=\bigcup_{\alpha <\omega_1} (\kappa_{2\alpha+1},\kappa_{2\alpha}]$... $\endgroup$ – Ergonvi Jan 16 '15 at 20:36
  • $\begingroup$ It doesn't work,for example then $\kappa_{\omega}$ should be in both of $X$ and $Y$. $\endgroup$ – Mohammad Golshani Jan 18 '15 at 4:43

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