Timeline for Is the supremum of L-definable cardinals silver-indiscernible
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 18, 2022 at 1:45 | vote | accept | Reflecting_Ordinal | ||
| Jun 3, 2021 at 17:16 | comment | added | Farmer S | Extending what @ElliotGlazer mentioned above, in fact the least $L$-indiscernible is exactly the set of all countable ordinals which are definable in $L$ from $V$-cardinals. | |
| Jun 3, 2021 at 16:56 | comment | added | Elliot Glazer | @Reflecting_Ordinal $i_0$ is the sup of countable ordinals which are definable in $L$ from $V$-cardinals (equivalently, from $\omega_n^V$'s). | |
| Jun 3, 2021 at 11:44 | vote | accept | Reflecting_Ordinal | ||
| May 18, 2022 at 1:45 | |||||
| Jun 3, 2021 at 11:19 | comment | added | Reflecting_Ordinal | Thank you, now I've understood it. Then, how could we understand how large the least Silver indiscernible is? For example, are there some "canonical" sequence of ordinal of length $\omega$ with limit the least Silver indiscernible? | |
| Jun 3, 2021 at 11:14 | comment | added | Joel David Hamkins | Because then $\alpha$ would also be fully correct, but $\kappa$ was the first fully correct ordinal. | |
| Jun 3, 2021 at 11:13 | comment | added | Reflecting_Ordinal | @Hamkins why can't $L_\kappa$ have a smaller $L_\alpha$ elementary substructure? | |
| Jun 3, 2021 at 10:57 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
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| Jun 3, 2021 at 10:51 | comment | added | Joel David Hamkins | @MonroeEskew You are right, and I have now edited to make a correct argument. | |
| Jun 3, 2021 at 10:51 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
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| Jun 3, 2021 at 10:26 | comment | added | Reflecting_Ordinal | Of course we should talk in the standard model, because in a nonstandard model the supremum may not exist. | |
| Jun 3, 2021 at 9:59 | comment | added | Monroe Eskew | If $\xi$ is a Silver indisernible, then it is inaccessible in $L$, so there are many $\alpha<\xi$ such that $L_\alpha \prec L_\xi$. Thus $L_\kappa \prec L$ does not imply $\kappa$ is an indiscernible. | |
| Jun 3, 2021 at 9:45 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
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| Jun 3, 2021 at 9:33 | history | answered | Joel David Hamkins | CC BY-SA 4.0 |