Talk:First uncountable ordinal

[0,ω₁) countably compact?

"The topological space [0,ω₁) is sequentially compact but not compact (nor even Lindelöf or countably compact). "

I think it is countably compact, and so it says in Order topology and in my topology reference. I'll go ahead and change it. YohanN7 (talk) 00:01, 21 July 2009 (UTC)

Done. Hope the conclusions about metrization aren't affected. YohanN7 (talk) 00:07, 21 July 2009 (UTC)

It's not metrizable because it's not Second Sountable. So i added "not second countable". Reason: All ordinal spaces satisfy all separarion axioms. Regular + Second Countable would imply Metrizable + Separable. YohanN7 (talk) 00:51, 21 July 2009 (UTC)

If ω₁+1 is perfectly normal then cof(ω₁)=ω. JumpDiscont (talk) 08:45, 6 June 2011 (UTC)

If the axiom of choice holds, then ω₁ is a regular ordinal, that is, its cofinality is not ω. Any continuous function from [0,ω₁] to the reals must be constant on [α,ω₁] for some α<ω₁. JRSpriggs (talk) 19:09, 6 June 2011 (UTC)

References

I added Jech's book as a reference. This is perhaps overkill, any book on Set Theory will define omega_1. (Kunen, Komjath, Just+Weese, Drake+Singh, Deiser, ...) Also some books in Algebra and/or Analysis will define omega_1 and at least mention some basic properties.

I also added "Counterexamples in topology" for the topological properties.

--Aleph4 (talk) 15:52, 23 May 2011 (UTC)

Compact if and only if Lindelof

"a countably compact space is compact if and only if it is Lindelöf" I believe that this is only true for T1 spaces, though I don't have a counterexample in mind. DanRaies (talk) 00:27, 5 April 2023 (UTC)

Nevermind, I was mistaken. "Countably compact" means every countable cover has a finite subcover. "Lindelof" means every cover has a countable subcover. "Compact" means every cover has a finite subcover. It is obvious that a countably compact space is compact if and only if Lindelof. DanRaies (talk) 22:33, 5 April 2023 (UTC)