Timeline for Can we iteratively reflect on self elementary embeddable stages of the cumulative hierarchy?
Current License: CC BY-SA 4.0
37 events
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| Jan 19, 2023 at 12:20 | comment | added | Zuhair Al-Johar | @FarmerS, your formula $\phi$ doesn't meet the specifications I've given for the use of $j $ in it (I've updated the question with those specifications), also you are double quantifying over $\alpha$ in a nested manner, which makes the axiom using it a non well formed formula. | |
| Jan 19, 2023 at 12:04 | comment | added | Zuhair Al-Johar | @Holo, ,Farmer sorry for the delay, but I'll need some time to read it and return back to you. | |
| Jan 19, 2023 at 0:21 | history | edited | Holo | CC BY-SA 4.0 |
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| Jan 18, 2023 at 23:50 | comment | added | Holo | @ZuhairAl-Johar I edited my answer, I hope it helps | |
| Jan 18, 2023 at 23:49 | history | edited | Holo | CC BY-SA 4.0 |
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| Jan 18, 2023 at 22:23 | comment | added | Farmer S | But that's not true, because $\mathrm{dom}(j_\alpha)=W_\alpha$ and $j_\alpha$ is elementary, so if we let $x\in W_\alpha$ be a non-function, then $j_\alpha(x)$ is not a function. So the theory is inconsistent. | |
| Jan 18, 2023 at 22:23 | comment | added | Farmer S | But then the following formula $\phi$ is true: "For every $\alpha\in\mathrm{dom}(j)$, $j(\alpha)$ is a function". So by the reflection scheme, we can find some $\alpha$ such that $\phi^{W_\alpha}$ holds. You also say that there is $\lambda$ such that $W_\alpha=V_\lambda$. So $\phi^{V_\lambda}$ holds. But $V_\lambda$ has no interpretation for the symbol $j$. I suppose what you really meant to say was that $\phi^{(W_\alpha,j_\alpha)}$ holds. Well, then $(W_\alpha,j_\alpha)\models$"For every $\beta\in\mathrm{dom}(j_\alpha)$, $j_\alpha(\beta)$ is a function"... | |
| Jan 18, 2023 at 22:22 | comment | added | Farmer S | You seem to be blurring the distinction between symbols and their interpretations. Okay, let's do that (you also do that at the outset by adding the "partial functions W,j to the language ZF"). But $\phi$ must be in (a subset of) the language of the original structure (because it appears as the hypothesis in the reflection scheme formula)... | |
| Jan 18, 2023 at 22:16 | comment | added | Holo | @ZuhairAl-Johar I will edit soon today (or tomorrow morning) to write a fully detailed explanation about all of the discussion in the comments inside of the post, and I hope it will clear things up | |
| Jan 18, 2023 at 22:13 | comment | added | Zuhair Al-Johar | I mean every occurrence of $j$ in the formula must be followed by $\alpha$ as subscript | |
| Jan 18, 2023 at 22:12 | comment | added | Farmer S | Do you mean that the language is set theory + j, and you want to interpret j with $j_\alpha$? | |
| Jan 18, 2023 at 22:10 | comment | added | Farmer S | $j_\alpha$ is not a symbol; it's a function in the model, no? | |
| Jan 18, 2023 at 22:10 | comment | added | Zuhair Al-Johar | @FarmerS, the language of set theory + $j_\alpha$. | |
| Jan 18, 2023 at 22:08 | comment | added | Farmer S | In the reflection scheme, what language is the formula $\phi$ in? | |
| Jan 18, 2023 at 22:07 | comment | added | Zuhair Al-Johar | @FarmerS, I don't get your objection. It's unlcear. | |
| Jan 18, 2023 at 22:01 | comment | added | Farmer S | @ZuhairAl-Johar it still doesn't work, and actually the reflection scheme doesn't even make sense, if you are intending the $\phi$ in the reflection scheme to be in the language with symbol $j$, since you write "$W_\alpha\models\phi$", but $W_\alpha=V_\lambda$ for some $\lambda$, which doesn't have any interpretation for $j$. | |
| Jan 18, 2023 at 21:34 | comment | added | Zuhair Al-Johar | @Holo, I think the problem was in the last axiom, which I've coined it to allow the use of $j_\alpha$ in the formula of reflection but I made the wrong restriction. I've corrected it. Perhaps, this corrects matters. | |
| Jan 18, 2023 at 21:05 | comment | added | Zuhair Al-Johar | @Holo, I'm sure you are right, I can feel it, but I can't fathom it. Replacement using $j_\alpha$ of course hold in the $W_\alpha$ world, there is no restriction on it. The critical point of $j_\alpha$ is ought to be a Reinhardt cardinal, there is something I'm missing. | |
| Jan 18, 2023 at 20:48 | comment | added | Holo | @ZuhairAl-Johar $W_{α+1}$ can see that the $κ$ such that $W_α=V_κ$ is $I3$, but for it to be Reinhardt you need to have that $W_α$ internally see the replacement of $j_{α}$ | |
| Jan 18, 2023 at 20:37 | comment | added | Zuhair Al-Johar | @Holo, but that can be seen in $W_{\alpha+1}$ about $W_\alpha$, or is it the case that I must allow $W_\lambda, j_\lambda$ symbols to be used in Elementarity axiom forall $\lambda < \alpha$, for that to be seen in it. | |
| Jan 18, 2023 at 20:29 | comment | added | Holo | @ZuhairAl-Johar If you want $(W_α,∈,j_α)$ to be a model of $ZF+Reinhardt$ you need $j_{α}^{(W_α,∈,j_α)}$ to be elementary embedding, there is no reason to believe that $j_α^{(W_α,∈,j_α)}$ have anything related to $j_α^V$. To maybe better understand this, try to think about $j_{2^{2^{2^{|α|}}}}^{(W_α,∈,j_α)}$, it can't be $j_{2^{2^{2^{|α|}}}}$ because $j_{2^{2^{2^{|α|}}}}$ is a function between $W_{2^{2^{2^{|α|}}}}$ to itself, but $W_α$ doesn't see anything close to $W_{2^{2^{2^{|α|}}}}$ | |
| Jan 18, 2023 at 20:21 | comment | added | Zuhair Al-Johar | @Holo, But $j_\alpha$ is an elementary embedding from $W_\alpha$ to $W_\alpha$ and $W_\alpha \models \sf ZF$, so it is an elementary embedding from a universe of disoucrse of ZF to itself, so how can there be no Reinhardt cardinal in $W_\alpha$? $j_\alpha$ is not trivial, so it must have a critical point, why that isn't a Reinhardt's cardinal? | |
| Jan 18, 2023 at 20:13 | comment | added | Holo | @HanulJeon no worries, with the additional context your comment is on point | |
| Jan 18, 2023 at 20:12 | comment | added | Hanul Jeon | I supposedly assume it is an elementary embedding over $W_\alpha$. It is my fault that I did not list every axiom I should state to avoid your confusion. | |
| Jan 18, 2023 at 20:11 | comment | added | Holo | @HanulJeon "separation and replacement for formulas with $j_α$" still doesn't mean $W_α$ will interpret $j_α$ as elementary embedding, I don't understand how separation and replacement for $j_α$ will do anything without any extra (non-ZF related) axioms | |
| Jan 18, 2023 at 20:06 | comment | added | Hanul Jeon | This is the reason why I stated "Separation and Replacement for formulas with $j$" (and in this case, schemas with $j_\alpha$ over $W_\alpha$.) | |
| Jan 18, 2023 at 20:03 | comment | added | Holo | @HanulJeon Requiring $\forall \alpha \,( W_\alpha \models \sf ZF)$ with the new language does not makes each $W_α$ satisfy $ZF+j_α\text{ is elementary embedding}$, $W_α$ may interpret $j_α$ different from how $V$ interpret it | |
| Jan 18, 2023 at 19:57 | comment | added | Hanul Jeon | Holo, I do not understand the point of your comment. What I pointed out was that Zuhair's theory is interpreted in your setup even if we additionally assume $W_\alpha$ models $\mathsf{ZF}$ over Zuhair's theory. If we require each $W_\alpha$ to satisfy $\mathsf{ZF}_j$ and $j$ is an elementary embedding from $V$ to $V$ itself, then the existence of $W_\alpha$ would demonstrate there is a transitive model of $\mathsf{ZF}$ with a Reinhardt cardinal. | |
| Jan 18, 2023 at 19:54 | comment | added | Holo | @ZuhairAl-Johar as FarmerS said, just requiring it being a model of $ZF$ doesn't add any power, you need to require something on how it interpret $j_α$ to change anything | |
| Jan 18, 2023 at 19:53 | comment | added | Holo | @HanulJeon it does not matter how to formulate it, to get Reinhardt you need "ZF+(schemas including j)+**j is elementary embedding**", just adding symbols without any restriction on the symbols doesn't change anything (let $L=FOST + F + R + C$ where $F$ are set of new function symbols, $R$-set of relations symbols, $C$ set of new constants, interpret every new symbol as $∅$, and any model of ZF will be a model of ZF with the new language with schemas allowing the new symbols) | |
| Jan 18, 2023 at 19:47 | comment | added | Farmer S | @ZuhairAl-Johar ZFC + I3 at $\alpha$ implies $V_\alpha\models$ ZFC, but $(V_\alpha,j)\not\models$ ZFC$_j$ where $j:V_\alpha\to V_\alpha$ witnesses I3 at $\alpha$. | |
| Jan 18, 2023 at 19:00 | comment | added | Zuhair Al-Johar | @HanulJeon, There is no restriction here on formulas used in Separation and replacement, so of course they can use $j$. | |
| Jan 18, 2023 at 18:55 | comment | added | Hanul Jeon | @Zuhair It depends on how to formulate $\mathsf{ZF}$. If we do not allow Separation and Replacement for formulas with $j$, then Holo's answer still works even if you assume $W_\alpha$ models $\mathsf{ZF}$. | |
| Jan 18, 2023 at 18:40 | comment | added | Zuhair Al-Johar | If one adds the axoim that $\forall \alpha \,( W_\alpha \models \sf ZF)$, then would that interpret $\sf ZF + Reinhardt $ at each $W_\alpha$? | |
| Jan 18, 2023 at 18:06 | comment | added | Holo | @ZuhairAl-Johar No, it is much weaker. The problem is that having a lot of elementary embedding doesn't give us a way to "stich them together" to get Reinhardt. And even if you had a way to stich them together there is no reason to believe that the "resulting stich" won't be the identity. (For example, it is consistent [relative to blah blah] that there is proper class of $I3$ ordinals, and for every ordinal $α$, there is some $β$ such that the elementary embedding for the $I3$ ordinals above $β$ are the identity on $V_α$, so the embeddings are "eventually the identity") | |
| Jan 18, 2023 at 17:56 | comment | added | Zuhair Al-Johar | I thought this theory is stronger than Reinhardt cardinals | |
| Jan 18, 2023 at 17:49 | history | answered | Holo | CC BY-SA 4.0 |