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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:)

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    The classical method, see The example 3.1.27 of Engelking's book.2012-06-25

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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.

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    @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.2012-06-25