Talk:First uncountable ordinal

[0,&omega;1) countably compact?
"The topological space [0,&omega;1) 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 &omega;1+1 is perfectly normal then cof(&omega;1)=&omega;. JumpDiscont (talk) 08:45, 6 June 2011 (UTC)


 * If the axiom of choice holds, then &omega;1 is a regular ordinal, that is, its cofinality is not &omega;. Any continuous function from [0,&omega;1] to the reals must be constant on [&alpha;,&omega;1] for some &alpha;<&omega;1. JRSpriggs (talk) 19:09, 6 June 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)