Timeline for Embeddability into $\beta\omega$ and $\omega^*$
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 11, 2022 at 11:36 | vote | accept | Damian Sobota | ||
| Mar 9, 2021 at 18:20 | answer | added | KP Hart | timeline score: 6 | |
| Dec 12, 2018 at 18:46 | comment | added | Will Brian | @DamianSobota: Good to know that $\mathfrak{c} = \aleph_2$ isn't strictly necessary. And you're correct that $\omega^*$ is not basically disconnected. But $\beta \omega$ is basically disconnected ("basically disconnected" is a weakening of "extremally disconnected") and $\beta \omega$ and $\omega^*$ embed in each other (which implies that a space embeds in $\omega^*$ if and only if it embeds in $\beta \omega$). | |
| Dec 12, 2018 at 17:41 | comment | added | Damian Sobota | @WillBrian: And at the end of the first section they write that the example can be constructed under MA + $\mathfrak{c}=\kappa^+$ for any regular uncountable cardinal. | |
| Dec 12, 2018 at 17:36 | comment | added | Damian Sobota | @WillBrian: so, Will, as I understand, it follows that $\omega^*$ is not basically disconnected, is it? | |
| Dec 12, 2018 at 17:16 | comment | added | Not Mike | @WillBrian Seems I misread the initial part of the question. Lol | |
| Dec 12, 2018 at 17:07 | comment | added | Will Brian | @AsafKaragila: Well, almost anyway -- I'd have to read through their proof (which I haven't) to know whether they really need $\mathfrak{c} = \aleph_2$ or not. For now we can at least say that stronger forcing axioms that give you $\mathfrak{c} = \aleph_2$ for free would disprove it outright. | |
| Dec 12, 2018 at 17:05 | comment | added | Asaf Karagila♦ | @NotMike: Make yourself Comfortable, why not? | |
| Dec 12, 2018 at 17:04 | comment | added | Asaf Karagila♦ | @Will: Thanks! I'd even say that MA is enough to disprove this, rather than "isn't enough to give you ..." | |
| Dec 12, 2018 at 17:03 | comment | added | Not Mike | With regards to the second question, I think there is something in Comfort's Theory Of Ultrafilters. But I'd have to check. | |
| Dec 12, 2018 at 17:03 | comment | added | Will Brian | @AsafKaragila: van Douwen and van Mill proved that $\mathsf{MA}+\mathfrak{c} = \aleph_2$ implies there is a compact $F$-space that does not embed in $\beta \omega$. So the answer to your comment is no, $\mathsf{MA}$ does not suffice to give you all spaces of weight at most $\mathfrak{c}$, and in fact it gives you the opposite. jstor.org/stable/1998148?seq=1#metadata_info_tab_contents | |
| Dec 12, 2018 at 16:50 | comment | added | Asaf Karagila♦ | I didn't say that you can assume. It's just that when you assume more you can usually prove more. So an obvious starting point would be with large cardinals. I think. | |
| Dec 12, 2018 at 16:49 | comment | added | Damian Sobota | What's going on under MA is a good question, I don't know. And concerning possible properties of $\kappa$, a priori I cannot assume anything. | |
| Dec 12, 2018 at 16:34 | comment | added | Asaf Karagila♦ | Obvious first quesiton: What happens under MA? Can you even get weight at most $\frak c$ in that case? Second obvious question: What happens in the third question when you consider some large cardinals (e.g. ones that admit lots of ultrafilters, like measurable cardinals) instead of just some arbitrary cardinal? | |
| Dec 12, 2018 at 15:46 | history | asked | Damian Sobota | CC BY-SA 4.0 |