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In my set theory class the professor said that if $\alpha$ is an ordinal of the form $\alpha=\kappa + \beta$, where $\kappa$ is an infinite cardinal and $\beta$ is an ordinal less than $\kappa$, then $[0,\alpha[$ is homeomorphic to $[0,\kappa[$.

I think that's no correct because $[0,\alpha[$ would be compact (because $\alpha$ is successor) and $[0,\kappa[$ no (because $\kappa$ is a limit ordinal). Could someone help me?

Thanks in advance.

Ergonvi
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  • To complement Brian's answer, one can state precisely when two ordinals are homeomorphic. See here. – Andrés E. Caicedo Jan 12 '15 at 20:05
  • @Andrés: Ah, nice one; that was before I started here, and I hadn’t previously run across it. – Brian M. Scott Jan 12 '15 at 20:11
  • @Brian Another interesting question is to classify those ordinals $\alpha$ such that if $\beta$ is homeomorphic to $\alpha$, then it contains a subset of type $\alpha$ and homeomorphic to $\alpha$ (Baumgartner would say that the subset is order-homeomorphic to $\alpha$). – Andrés E. Caicedo Jan 12 '15 at 20:16
  • The downvote on this question is hard to construe as anything but a display of ignorance: it’s an intelligent question with an obviously adequate context. The only other plausible explanation is vandalism for the sheer pleasure of it. – Brian M. Scott Jan 12 '15 at 21:14

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You are correct in thinking that the professor is mistaken, but it’s not true that $\alpha$ is necessarily a successor. What is true is that $[0,\alpha)$ is homeomorphic to the disjoint union of $[0,\kappa]$ and $[0,\beta)$, which is compact if and only if $\beta$ is a successor ordinal. If $\beta$ is a successor ordinal, $[0,\alpha)$ is homeomorphic to the ordinal sum $\beta+\kappa+1=\kappa+1$ and hence to $[0,\kappa]$, while $[0,\kappa)$ is not compact.

Brian M. Scott
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