Timeline for Where do the universe embedds to in Reinhardt's cardinals setting?
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19 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 23, 2022 at 12:04 | comment | added | Zuhair Al-Johar | Ah! I see, $j$ is not surjective! So, there can exist $\alpha \in j(\lambda)$ where $\alpha$ is outside the range of $j $ | |
| Dec 23, 2022 at 9:14 | comment | added | Asaf Karagila♦ | Why does that entail that? | |
| Dec 23, 2022 at 8:13 | comment | added | Zuhair Al-Johar | but $j$, being an elementary embedding, is supposed to preserve $\in$, that is $x \in y \iff j(x) \in j(y)$, this entails that for ANY set $s$ we have $j(s)=j[s]$ | |
| Dec 23, 2022 at 7:42 | comment | added | Asaf Karagila♦ | Since anything above the critical point will not satisfy $j(\lambda) = j[\lambda]$, I don't see why this is an issue. In fact you don't necessarily have $\sup j[\lambda] = j(\lambda)$... | |
| Dec 23, 2022 at 6:55 | comment | added | Zuhair Al-Johar | I always thought that for any elementary embedding $j$ we have $j(x)=j[x]$ for all $x$, since elementary embeddings must preserve membership $\in$. Now, let $\lambda$ be above the critical point of $j$, and $\lambda < \kappa$, we have $V_\kappa \models ordinal (\lambda)$, now if $j(\lambda)= j[\lambda]$, then by your argument $ V_\kappa \models \neg ordinal(j(\lambda))$, violating $j$ be an elementary embedding? | |
| Dec 21, 2022 at 17:52 | vote | accept | Zuhair Al-Johar | ||
| Dec 21, 2022 at 17:37 | comment | added | Asaf Karagila♦ | Because then $j^{n+1}(\alpha)<j^n(\alpha)$ and you got yourself a decreasing sequence of ordinals. Normally, something that is frowned upon in the context of ZF and related theories. Note that if $j$ is at all amenable to the universe, then this gives you a fairly simple definition as far as Separation goes. | |
| Dec 21, 2022 at 17:07 | comment | added | Zuhair Al-Johar | Why not $j(\alpha) <\alpha$? | |
| Dec 21, 2022 at 15:45 | comment | added | Asaf Karagila♦ | No. If $j$ is non-trivial, there is some $\alpha<\kappa$ such that $j(\alpha)>\alpha$. Then the first such $\alpha$ will not be in $j[\kappa]$. Full stop. | |
| Dec 21, 2022 at 15:33 | comment | added | Zuhair Al-Johar | Of course the first. Still $j^+(\kappa)$ is a transitive set of transitive sets, and so an ordinal. I'm still not seeing how $j^+(\kappa)$ fails to be an ordinal? | |
| Dec 21, 2022 at 15:10 | comment | added | Asaf Karagila♦ | Either the critical point of $j$ is below $\kappa$, in which case $j[\kappa]\neq\kappa$, or it is trivial. Those are the only two options. Choose one of them, then go back to my previous comments. | |
| Dec 21, 2022 at 14:50 | comment | added | Zuhair Al-Johar | No! I didn't make that assumption. | |
| Dec 21, 2022 at 14:22 | comment | added | Asaf Karagila♦ | You are assuming that $\kappa$ was the critical point of the embedding? In that case $j\colon V_\kappa\to V_\kappa$ was the identity. Again, your mistakes stem from an overloading and overusing the letter $\kappa$. | |
| Dec 21, 2022 at 14:10 | comment | added | Zuhair Al-Johar | $j^+(x)=j(x)$ for all $x \in V_\kappa$, so all ordinals in $\kappa$ will be sent by $j^+$ to ordinals in $\kappa$, now $j^+$ would preserve transitivity, and so $j^+(\kappa)$ would be a transitive set of transitive sets, and this is an ordinal in a well founded milieu. | |
| Dec 21, 2022 at 10:17 | comment | added | Asaf Karagila♦ | Why would it be an elementary embedding? $j^+(\kappa)=j[\kappa]$ is not an ordinal if $j$ is non-trivial. This may preserve some modicum of elementarity with regards to the second-order theory of $V_\kappa$, but it is no longer a fully elementary embedding. There is a whole theory on when and how these embeddings can be lifted and by how much. It is complicated, it is difficult, and it is full of subtlety which you are currently sweeping off the table. I am not an expert on those things either, but Gabe and Farmer both have papers on this and both are around often. | |
| Dec 21, 2022 at 10:14 | comment | added | Zuhair Al-Johar | Good! Try to go one level above MK, that is let us work with ZF + some strong inaccessible exists. Now take $V_\kappa$ where $\kappa$ is the first strong inaccessible. Let $j: V_\kappa \to V_\kappa$, and such that $j$ is an elementary embedding that is non-trivial; now take the function $j^+ : V_{\kappa +1} \to V_{\kappa+1}; j^+(x)= j [x]=\{j(y) \mid y \in x\}$, this would also be elementary embedding. Now, what is $j^+(\kappa)$ ? If it equals $\kappa$, then $j$ would be an automorphism which can't be, so $j^+(\kappa) < \kappa$ but this embeds $V_\kappa$ to an element of it? | |
| Dec 21, 2022 at 9:55 | comment | added | Asaf Karagila♦ | Yes, but why would it be surjective, unless it is trivial? | |
| Dec 21, 2022 at 9:44 | comment | added | Zuhair Al-Johar | if an elementary embedding is also surjective wouldn't that make it an automorphism? | |
| Dec 21, 2022 at 8:57 | history | answered | Asaf Karagila♦ | CC BY-SA 4.0 |