Timeline for Can reflection in $V$ and reflection out of $V$ principles be used in the same theory?
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36 events
| when toggle format | what | by | license | comment | |
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| Jun 9, 2019 at 19:02 | comment | added | Zuhair Al-Johar | this is the stronger one: mathoverflow.net/questions/319319/… | |
| Jun 9, 2019 at 18:36 | comment | added | Zuhair Al-Johar | YES! correct. I think this is right. It is not very strong. The strong theory is another one, not this. | |
| Jun 9, 2019 at 18:33 | comment | added | Master | However, isn't this theory not very strong. Consider the mode $V_\kappa$, where $\kappa$ is inaccessible. Any $V_\alpha\prec V_\kappa$ would satisfy counter reflection and reflection, and so this is theory is consistent relative to large cardinals? | |
| Jun 9, 2019 at 18:24 | comment | added | Zuhair Al-Johar | @Master, EXACTLY! | |
| Jun 9, 2019 at 16:02 | comment | added | Master | I see. So for example, $\forall x(\forall y(\phi))^V\rightarrow \forall x(\forall y(\phi))$, which is one of the advantages of counter reflection? | |
| Jun 9, 2019 at 13:25 | comment | added | Zuhair Al-Johar | @Master, $\exists x\in V(\phi)\leftrightarrow \exists x(\phi)$ only when $\phi$ has the needed qualifications, this will cut the induction flow when $V$ is included in $\phi$. So this won't work. | |
| Jun 8, 2019 at 21:38 | comment | added | Zuhair Al-Johar | @Master, Ah! I see what you want to do. I need to think about it. Its too late at night now. I'll continue this tomorrow! | |
| Jun 8, 2019 at 21:31 | comment | added | Master | $(x\in y^V)$ is the sentence $(x\in y)$ with all its quantifiers bounded; as it has no quantifiers, it is identical to $(x\in y)$ . What I am trying to show is that for every formula $\phi$, $\phi^V\leftrightarrow \phi$. I am trying to do this by induction: First for atoms, and then the connectives and negation, and then the existential quantifier. | |
| Jun 8, 2019 at 21:26 | comment | added | Zuhair Al-Johar | @Master, what is $(x \in y)^V$, the way how I used $\phi^V$ is to mean $\phi$ is a "sentence" where its quantifiers are bound by $V$, the way how I see it is that $x \in y$ is not a sentence? it contains two free variables. | |
| Jun 8, 2019 at 21:12 | comment | added | Master | We can prove by induction $\phi^V\leftrightarrow \phi$ for every formula $\phi$. It is clear that $(x\in y)^V\leftrightarrow (x\in y)$, and if $\phi$,$\psi$ are absolute so is $\lnot\phi$ and $\phi\land \psi$. Then if $\phi$ is absolute, $\exists x\in V(\phi)\rightarrow \exists x(\phi)$ and by reflection $\exists x(\phi)\rightarrow \exists x\in V(\phi)$, and so $\exists x(\phi)\leftrightarrow \exists x(\phi)^V$. | |
| Jun 8, 2019 at 21:07 | comment | added | Zuhair Al-Johar | @Master, I don't know what do you mean when you say "by induction"? how do you use "induction" to generalize what is inside $V$ to go beyond $V$? I used counter-reflection to do that, but apparently you have something else in mind. | |
| Jun 8, 2019 at 20:53 | comment | added | Master | Given any predicate $\phi(x)$, $V\vDash (\lnot\exists X\lnot(\exists Y(Y=\{x\in X|\phi(x)\})))$. Now, by induction we can see that then the universe $W=\{x|x=x\}$ satisfies this, and we do not even have to use reflection out of $V$ to see this, and so separation is satisfied. | |
| Jun 8, 2019 at 20:32 | comment | added | Zuhair Al-Johar | @Master, No! class comprehension only proves separation over the class $V$, but it doesn't prove separation over higher classes, for example separation over the class $P(V)$ or its power, etc... | |
| Jun 8, 2019 at 20:20 | comment | added | Master | Doesn't class comprehension proof class separation? I.e. $\phi(x)\leftrightarrow x\in Z\land \psi(x)$? | |
| Jun 8, 2019 at 20:00 | comment | added | Zuhair Al-Johar | @Miller's separation over classes. That was the main point behind counter-reflection. | |
| Jun 8, 2019 at 19:38 | comment | added | Master | Well I still don't see what you could derive from counter-reflection that you can't from reflection, so I don't see the point of this question, unless I am missing something really obvious. | |
| Jun 8, 2019 at 19:36 | comment | added | Zuhair Al-Johar | @Master, I don't think so, anyhow I'm not sure. | |
| Jun 8, 2019 at 18:04 | comment | added | Master | Yes, it seems counter reflection is just reflection from a different perspective. Kind of like the schema $\forall x\in V(\phi(x,x_0...x_n))\rightarrow \forall x(\phi(x,x_0...x_n))$ is the same as $\exists x(\phi(x,x_0...x_n))\rightarrow \exists x\in V(\phi(x,x_0...x_n))$. There both absoluteness from a different direction. | |
| Jun 8, 2019 at 17:47 | comment | added | Zuhair Al-Johar | are you saying that counter-reflection is redundant? | |
| Jun 8, 2019 at 17:41 | comment | added | Master | Wait, but if every $FOL(=,\in)$ formulas is reflected by refection, and the same is true by counter reflection, then what is the difference? | |
| Jun 8, 2019 at 16:55 | comment | added | Zuhair Al-Johar | @yes that's what I mean. as far as counter-reflection is concerned. | |
| Jun 8, 2019 at 16:50 | comment | added | Master | Both reflection and counter-reflection are restricted to $L(=,\in)$. "if $\varphi(y, x_1,..,x_n)$ is a formula in $FOL(=,\in)$","if $\varphi$ is a sentence in $FOL(=,\in)$, i.e. doesn't use the symbol $V$." | |
| Jun 8, 2019 at 16:48 | comment | added | Zuhair Al-Johar | @Master, I meant the part that $\varphi$ doesn't use the symbol $V$. | |
| Jun 8, 2019 at 16:37 | comment | added | Master | You mean the part it says $\forall x_0...x_n\in V$? Doesn't all that mean is parameters are restricted to $V$? | |
| Jun 8, 2019 at 3:41 | comment | added | Zuhair Al-Johar | @Master, there is a restriction on $V$ as mentioned in the axiom, so its not $\phi^V$ for whatever $\phi$. | |
| Jun 7, 2019 at 21:31 | comment | added | Master | I am confused. The reflection schema states roughly $V\prec W$, where $W=\{x|x=x\}$. As a consequence, $\phi^V\rightarrow \phi$, unless I have mistaken something. | |
| Dec 18, 2018 at 18:35 | comment | added | Zuhair Al-Johar | @MonroeEskew so from these comments of yours and Hamkins referred posts I understand that the theories I'm referring to here (containing reflection in and reflection out schemes on top of the other axioms) are all consistent! but anyhow the axiomatization that I've presented here is by far much less complex than the ones you (and Hamkins) are expressing. This is in itself interesting, at least to me. | |
| Dec 18, 2018 at 18:26 | comment | added | Monroe Eskew | Yep! Mahlos are everywhere. | |
| Dec 18, 2018 at 18:20 | comment | added | Zuhair Al-Johar | @MonroeEskew do your comments apply to adding the reflection out schema on top of the referred theory in the post that has the limitation of size axiom? this has a mahlo cardinal as its model, (see link in the head post), so is this also mild in your opinion? | |
| Dec 18, 2018 at 14:48 | comment | added | Joel David Hamkins | See mathoverflow.net/a/103779/1946 for more discussion. Also, this theory arises in many other questions on MO. See mathoverflow.net/search?q=user%3A1946+Feferman+theory. | |
| Dec 18, 2018 at 14:47 | comment | added | Monroe Eskew | Fix some $V_\alpha$ satisfying ZFC and pretend it is $V$ and the higher-rank sets are hyperclasses. Possibly $V_\alpha \prec V$. In such a situation, we have reflection out of $V$. Also everything true in $V$ reflects to $V_\alpha$. We can also have this on a club of $\beta$, so that also things can reflect below $\alpha$ simultaenously. | |
| Dec 18, 2018 at 14:06 | comment | added | Zuhair Al-Johar | @MonroeEskew your point is not clear? what do you want to say? | |
| Dec 18, 2018 at 11:57 | comment | added | Monroe Eskew | What about a schema with a constant symbol $c$ asserting that $c$ is some $V_\alpha$, and $V_\alpha$ is elementary in $V$? This is a mild theory. | |
| Dec 18, 2018 at 11:39 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
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| Dec 18, 2018 at 11:33 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
added 96 characters in body; edited title
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| Dec 17, 2018 at 19:26 | history | asked | Zuhair Al-Johar | CC BY-SA 4.0 |