We are familar with that for the first uncountable cardinality $\omega_1$, the topological space $[0,\omega_1]$ is compact. I find the proof for the $\omega_1$, is also for every regular cardinality. So is there a result for every regular cardinality, i.e.,

For every regular cardinality $\kappa$ with order topology, then $\kappa\cup \{\kappa\}$ is compact?

Am I right? Thanks for any help:)

  • $\begingroup$ What proof did you have in mind? $\endgroup$ – tomasz Jun 25 '12 at 0:19
  • $\begingroup$ @John Actually, The topological space $[0,\alpha]$ is compact for all ordinal $\alpha$ and the space $[0,\alpha)$ is never compact for limit $\alpha$. $\endgroup$ – azarel Jun 25 '12 at 0:22
  • $\begingroup$ The classical method, see The example 3.1.27 of Engelking's book. $\endgroup$ – Paul Jun 25 '12 at 0:23

For limit ordinal $\alpha$, $\alpha$ with order topology is not compact, since it is the strictly increasing union of $\beta<\alpha$, all of which are open (as subsets of $\alpha$).

On the other hand, any $\alpha+1$ is compact when equipped with order topology, since if we choose any family of open sets covering $\alpha+1$, we can find a subcover the size of which is the length of a descending sequence of ordinals (the elements of the sequence are maximal elements not covered by the intervals chosen so far, it can be done because any noninitial open interval has a maximal preceding element), so because any such sequence is finite, we have compactness.

Whether the number is limit, regular or cardinal does not matter. Any successor has that property.

  • $\begingroup$ Mmm, nice. Thanks tomasz. From your result, for every ordinal $\alpha$, $\alpha$ is countably compact? $\endgroup$ – Paul Jun 25 '12 at 0:48
  • $\begingroup$ No, if cofinality of $\alpha$ is $\omega$, for example (e.g. $\alpha=\omega$), $\alpha$ is the union of some $\omega$ numbers less than it, and we cannot hope to choose a finite subcollection that would still cover $\alpha$. If cofinality of $\alpha$ is greater than $\omega$, then I think $\alpha$ is countably compact. $\endgroup$ – tomasz Jun 25 '12 at 13:00
  • $\begingroup$ I may be pulling it a little, but I believe that for limit $\alpha$, and a given cardinal $\kappa$, we can choose a finite subcover of $\alpha$ from any cover of cardinality $\kappa$ if and only if $\kappa<cf(\alpha)$ $\endgroup$ – tomasz Jun 25 '12 at 13:06
  • $\begingroup$ @tomasz: Yes, that’s correct. If you’ve an open cover of cardinality less than $\operatorname{cf}\alpha$, some member of it must cover a tail of $\alpha$, and what remains is compact. $\endgroup$ – Brian M. Scott Jun 25 '12 at 14:17

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