Timeline for continuity points of elementary embeddings from $0^\sharp$
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 28, 2018 at 7:43 | vote | accept | Monroe Eskew | ||
| Apr 27, 2018 at 18:34 | answer | added | Gabe Goldberg | timeline score: 9 | |
| Apr 27, 2018 at 6:15 | history | edited | Monroe Eskew | CC BY-SA 3.0 |
added 43 characters in body
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| Apr 27, 2018 at 5:54 | history | edited | Monroe Eskew | CC BY-SA 3.0 |
added 141 characters in body
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| Apr 26, 2018 at 15:25 | comment | added | Victoria Gitman | Great! Thanks very much for the explanation! | |
| Apr 26, 2018 at 15:13 | comment | added | Gabe Goldberg | More generally, the elementary embeddings of $L$ are parameterized by the order embeddings of the Silver indiscernibles. See for example Schindler's Set Theory, Corollary 10.44 (5). | |
| Apr 26, 2018 at 15:05 | comment | added | Gabe Goldberg | The image of the Silver indiscernibles under $j'$ is a club class of order indiscernibles for $L$ and hence is contained in the Silver indiscernibles. | |
| Apr 26, 2018 at 14:31 | comment | added | Victoria Gitman | Almost got it. Why is $j'(\alpha_0)$ indiscernible? | |
| Apr 26, 2018 at 13:56 | comment | added | Gabe Goldberg | Derive the ultrapower embedding $j': L \to L$ from $j$ using $\alpha_0$ as a seed. Note that $j' \leq j$ pointwise on the ordinals since $j'$ factors into $j$. But $j$ is the pointwise minimum order-preserving function on the indiscernibles that moves $\alpha_0$. So $j'$ and $j$ agree on the indiscernibles. The indiscernibles generate $L$, so $j' = j$. | |
| Apr 26, 2018 at 13:04 | comment | added | Victoria Gitman | I don't see why $j$ has to be an ultrapower embedding. Is this obvious? | |
| Apr 26, 2018 at 7:37 | comment | added | Stamatis Dimopoulos | I don't have much experience with $0^{\#}$ but it may be relevant that usually this property is associated with strong compactness at $\delta$, and there are no inner models (yet) for such large cardinals. | |
| Apr 25, 2018 at 16:49 | comment | added | Gabe Goldberg | $j$ is an ultrapower embedding by an $L$-$\alpha_0$-complete $L$-ultrafilter on $\alpha_0$ using functions in $L$ so it is continuous at all ordinals whose $L$-cofinality is not $\alpha_0$. | |
| Apr 25, 2018 at 15:47 | history | asked | Monroe Eskew | CC BY-SA 3.0 |