Timeline for When is the generalized Cantor space $\kappa$-compact?
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Mar 31, 2016 at 11:36 | history | edited | Boaz Tsaban | CC BY-SA 3.0 |
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| Mar 31, 2016 at 11:14 | history | edited | Boaz Tsaban | CC BY-SA 3.0 |
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| Mar 8, 2016 at 19:58 | vote | accept | Boaz Tsaban | ||
| Mar 7, 2016 at 18:00 | comment | added | godelian | Boaz: Yes, I was applying the tree property, but I see that the branch might be covered. And would only have worked for one implication anyway. | |
| Mar 7, 2016 at 17:48 | answer | added | Joseph Van Name | timeline score: 10 | |
| Mar 7, 2016 at 17:45 | comment | added | Asaf Karagila♦ | Boaz: Looks like something I should have noticed. So I will unhedge my bet, and go with weakly compact. @godelian: I'm afraid I don't have the mental capacity right now to think about this sort of mathematics. | |
| Mar 7, 2016 at 17:44 | comment | added | Boaz Tsaban | @godelian: You mean you are applying the tree property to the tree induced by all these $\sigma$-s? How do you prove the branch is not covered? In any case, I think I see why the tree property suffices, but with a slightly more involved argument. And what about the converse implication, why is it necessary? | |
| Mar 7, 2016 at 17:36 | comment | added | Boaz Tsaban | @AsafKaragila: Be healthy, we need you here. :) Note that every open cover of $2^\kappa$ is refined by one with basic open sets. We assume $2^{<\kappa}=\kappa$, so we may restrict attention to open covers of cardinality $\kappa$. | |
| Mar 7, 2016 at 17:14 | comment | added | godelian | @Asaf: what about this argument by contradiction; for each $\alpha<\kappa$ the basic opens $[\sigma]$ with $\sigma \in 2^{\alpha}$ is an open covering of cardinality less than $\kappa$, so for at least one $\sigma$ the open $[\sigma]$ is not contained in the original open covering. Then by weak compactness there is a cofinal branch made of these $\sigma$'s which cannot be covered by the original open cover. | |
| Mar 7, 2016 at 16:59 | comment | added | Asaf Karagila♦ | If my memory serves me right this is strongly compact cardinals (I'd say weakly compact, but I am going to hedge my bets here, and claim my memory says in the weakly compact case it is only true for covers of cardinality $\kappa$). But I've got a mild case of the flu, so there's no reason to trust my memory. | |
| Mar 7, 2016 at 16:55 | history | asked | Boaz Tsaban | CC BY-SA 3.0 |