Timeline for What is the consistency strength of "Singular worldly that is inaccessible in an inner model"?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 26 at 21:05 | vote | accept | Asaf Karagila♦ | ||
| Aug 26 at 20:55 | answer | added | Farmer S | timeline score: 9 | |
| Aug 24 at 13:13 | comment | added | Andrés E. Caicedo | So, "$0^\sharp$ exists, and there is a wordly cardinal". | |
| Aug 24 at 12:03 | comment | added | Asaf Karagila♦ | Ahh, okay. I see what you mean now. Yes, I had meant sharps stronger than $0^\#$, but it seems that's not going to be the case necessarily. | |
| Aug 24 at 11:42 | comment | added | Farmer S | My main query was what is meant by "higher sharp", and then I was just pointing out that if the hypothesis is consistent, then it's also consistent with "$V=L[0^\sharp]$", so, for example, it doesn't prove that $(0^\sharp)^\sharp$ exists. | |
| Aug 24 at 11:05 | comment | added | Asaf Karagila♦ | Even more confused now... $V_\eta^{L[0^\#]}\models V=L[0^\#]$... Right? And so it is a model of $\sf ZFC$ in which there are no worldly cardinals. | |
| Aug 24 at 10:48 | comment | added | Farmer S | I'm not working in $V_\eta^{L[0^\sharp]}$. | |
| Aug 24 at 10:44 | comment | added | Asaf Karagila♦ | I am confused by your comment. If $\eta$ is the least worldly cardinal in $L[0^\#]$, working in $V_\eta^{L[0^\#]}$, there is no $\alpha$ for which $V_\alpha\models\sf ZF$, but $0^\#$ still exists. | |
| Aug 24 at 10:40 | comment | added | Farmer S | What do you mean by "higher sharp"? If $0^\sharp$ exists and there exists a singular cardinal $\eta$ such that $V_\eta\models$ ZF and $\eta$ is inaccessible in $L$, then the same hypothesis is true in $L[0^\sharp]$. (In fact, if $0^\sharp$ exists and $V_\eta\models$ ZF, then $\eta$ is inaccessible in $L$, and if $\eta$ is least such that $V_\eta\models$ ZF, then $\eta$ has cofinality $\omega$.) | |
| Aug 24 at 10:00 | history | asked | Asaf Karagila♦ | CC BY-SA 4.0 |