Timeline for Sequences of sequences of sequences and elementary embeddings
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Jan 21, 2021 at 10:28 | comment | added | Mohammad Golshani | What about the following idea: Suppose I have $2^\alpha=\alpha^{++}$ for inaccessibles below $\kappa$ but $2^\kappa=\kappa^+$ and suppose i define $F_\alpha$ in a reasonable way so that for each $\alpha < \kappa$ it takes the first $\alpha^+$ functions, but at $\kappa$ it enumerates all such functions (you described). In the sense of $M$, $G_\kappa$ can not enumerate all functions, so it is different from $F_\kappa$. | |
| Jan 14, 2021 at 17:01 | comment | added | Monroe Eskew | Perhaps one coherence property would be that this method of just putting $F_\kappa = j(\mathcal F \restriction \kappa)(\kappa)$ is reflected on a large set contained in $\kappa$. For example, suppose $j$ is a sufficiently strong embedding so that its derived ultrafilter is in $M$, and moreover this situation is even visible in $M$ by looking at a restricted extender which also lives in $M$. So at many $\alpha$, $F_\alpha$ is generated in this way, which means $F_\kappa$ is as well. Not sure if this helps. | |
| Jan 14, 2021 at 16:23 | comment | added | Asaf Karagila♦ | So can you ensure equality under any embedding by a coherence requirement on $\cal F$, rather than a definability requirement? | |
| Jan 14, 2021 at 15:56 | comment | added | Monroe Eskew | Actually this very issue came up when I was working with Yair on Radin forcing with interleaved posets and guiding generics. We wanted coherence, so we guaranteed equality by what I said above. | |
| Jan 14, 2021 at 15:54 | comment | added | Monroe Eskew | Well, we have examples showing that a given $\mathcal F$ can go either way. On the other hand, if it follows some definition that is absolute between $V$ and $M$, (like in 1.), then there is more constraint. | |
| Jan 14, 2021 at 15:52 | comment | added | Asaf Karagila♦ | Sure. But suppose that $\cal F$ is given, or that there are several different $j$s to consider in the overarching situation. (That being said, your suggestion does give a way to solve a particular problem, but I'm interested in the general case as well of what can we say about it under certain assumptions.) | |
| Jan 14, 2021 at 15:50 | comment | added | Monroe Eskew | One way of guaranteeing equality is to start with $\mathcal F \restriction \kappa$ and define $F_\kappa = j(\mathcal F \restriction \kappa)(\kappa)$. This also gives a way of guaranteeing inequality. | |
| Jan 14, 2021 at 15:22 | history | edited | Asaf Karagila♦ | CC BY-SA 4.0 |
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| Jan 14, 2021 at 13:59 | history | asked | Asaf Karagila♦ | CC BY-SA 4.0 |