Timeline for Can there be a segment of regular cardinals with the tree property capped by an almost-strongly-compact?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 5, 2018 at 21:40 | comment | added | Asaf Karagila♦ | Tim, it's a theorem of Specker, if my memory serves me right, that under the assumption that $\kappa^{<\kappa}=\kappa$, then $\kappa^+$ does not have a tree property. I think you can find the proof in Kanamori, but I'm not sure. The construction is somehow similar to the standard Aronszajn tree, but of course slightly different. | |
| Aug 5, 2018 at 20:25 | comment | added | Tim Campion | @YairHayut Actually -- where does $\rho < \kappa$ come from? | |
| Aug 5, 2018 at 20:15 | comment | added | Tim Campion | @YairHayut Thanks, I was not aware that the tree property definitely fails at the successor of any strongly inaccessible $\rho$ -- I thought this was something very peculiar to $\omega_1$. I would happily accept this as an answer. Is it easy to generalize standard constructions of an Aronszajn tree on $\omega_1$ to $\rho^+$? | |
| Aug 5, 2018 at 20:06 | comment | added | Yair Hayut | I think that this is still impossible (assuming $\mu^+ < \kappa$). Since $\kappa$ is $(\mu,\infty)$-strongly compact, there is a measurable cardinal $\mu \leq \zeta \leq \kappa$. In particular, there is a strongly inaccessible cardinal, $\rho$, in the half-open interval $[\mu, \kappa)$. In particular, $\rho^{<\rho} = \rho$ and the tree property will fail at $\rho^+$. | |
| Aug 5, 2018 at 16:02 | history | edited | Tim Campion | CC BY-SA 4.0 |
added 101 characters in body
|
| Aug 5, 2018 at 14:42 | history | edited | Tim Campion | CC BY-SA 4.0 |
added 223 characters in body; edited title
|
| Aug 5, 2018 at 14:32 | comment | added | Tim Campion | @YairHayut I wouldn't be surprised if I am missing something very, very basic. At the risk of being a bit ridiculous, I might change my question again. | |
| Aug 5, 2018 at 14:32 | comment | added | Yair Hayut | Do you know if a double successor cardinal can be almost strongly compact at all? I think that it would mean that there is a non-principle ultrafilter with completeness which is not a measurable cardinal, which is impossible. | |
| Aug 5, 2018 at 13:57 | comment | added | Tim Campion | @AsafKaragila Thanks, I realized that I was not asking what I intended to ask. I think you're right that clearly if $\mu$ is strongly compact then it has the tree property and $\mu^+$ is almost strongly compact. It turns out I had asked the wrong question! | |
| Aug 5, 2018 at 13:55 | history | edited | Tim Campion | CC BY-SA 4.0 |
added 30 characters in body; edited title
|
| Aug 5, 2018 at 7:57 | comment | added | Asaf Karagila♦ | What fails when $\mu$ is strongly compact? | |
| Aug 5, 2018 at 5:41 | history | edited | Tim Campion | CC BY-SA 4.0 |
added 53 characters in body
|
| Aug 5, 2018 at 4:31 | history | asked | Tim Campion | CC BY-SA 4.0 |