Timeline for Is Ackermann's set theory minus class comprehension equal to ZF?
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48 events
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| Feb 13, 2021 at 7:44 | vote | accept | Zuhair Al-Johar | ||
| Mar 11, 2020 at 13:46 | comment | added | Zuhair Al-Johar | I'm not understanding your answer. You need to explicitly say if the answer is YES or NO. Is there a real need for the class comprehension axiom of Ackermann or not? | |
| Jun 13, 2019 at 22:08 | comment | added | Master | Yes, I meant $\omega\cap V$. | |
| Jun 13, 2019 at 16:10 | comment | added | Zuhair Al-Johar | do you mean $\omega \cap V$ is a set, which is, since it is $\omega$ itself. The proof that $\omega$ is a set i.e. it is an element of $V$ is to define natural as an element of every class that has the empty set among its elements and that is closed under adjunction. Now clearly $V$ is one of those classes, so this mean that all naturals must be in $V$. Its more of a trick really. Clearly you cannot assume that $\omega \cup V$ is a set, since this is $V$ itself! and $V$ is a class but it is not a set, since a set is defined as what is an element of $V$ and we have $V \not \in V$ | |
| Jun 13, 2019 at 15:19 | comment | added | Master | I see, you are right. I was immediately assuming that $\omega\cup V$ was a set. I see now the flaw. | |
| Jun 13, 2019 at 5:01 | comment | added | Zuhair Al-Johar | try to prove the induction theorem for all formulas in this theory, you'll see you cannot. However you can prove induction and transfinite induction for formulas not using $V$ (from parameters in $V$). I'm raising this point just to make sure that non of your additional arguments are using induction over formulas using $V$, which would render them ineffective. | |
| Jun 13, 2019 at 4:52 | comment | added | Zuhair Al-Johar | induction in this theory only work for properties definable after a formula that doesn't use the symbol $V$. You can easily suppose the existence of a natural $n \not \in V$, how you'll prove in this theory that this is contradictory. Ordinarily we take the set of all ordinals smaller than $n$ that doesn't fulfill $P$ as a proof of violation of well foundedness of $n$, but to do that you need to use reflection and since $P$ here is the predicate of being an element of $V$, this is forbidden from being used. You see! | |
| Jun 13, 2019 at 4:50 | comment | added | Master | You can use proofs using any symbol in Ackermann set theory. You can just only use $L(\in,=)$ in reflection. But the language of $A$ is $L(\in,=,V)$. | |
| Jun 13, 2019 at 4:26 | comment | added | Zuhair Al-Johar | the point is that you cannot prove $\omega \subseteq V$ by induction. Simply induction for $V$ is not a theorem of this theory because you cannot use $V$. You cannot prove by induction that every natural an element of $V$. However $\omega \subseteq V$ is indeed a theorem of this theory, no doubt, but this is not proved by induction. | |
| Jun 13, 2019 at 1:00 | comment | added | Master | You prove by induction $\omega\subseteq V$. You don't define $\omega$ by induction. | |
| Jun 12, 2019 at 21:01 | comment | added | Zuhair Al-Johar | even if $0 \in V$ and for every natural $n \in V$ we have $n+1 \in V$ still that doesn't mean that all naturals are in $V$, you cannot use the predicate $`` \in V"$ in induction, because simply its forbidden in the reflection axiom. For example you can suppose that there is a natural [defined as a finite von Neumann ordinal] that is not in $V$, without reaching into any clear inconsistency by that argument alone. However, if you define natural as an element of the intersection class of all inductive classes, then yes here you can prove that every natural is in $V$. | |
| Jun 12, 2019 at 16:27 | comment | added | Master | What I mean is that $0\in V$, and $n+1=\{x|x\in n\lor x=n\}$, which is definable and so in $V$. And as $n+1^V=n+1$, $\omega\subseteq V$ and morover it is definable without using the symbol $V$, and $\omega^V=\omega$ and so $V\vDash Infinity$. | |
| Jun 12, 2019 at 13:37 | comment | added | Zuhair Al-Johar | There is some point! You say inductively we can prove $n \in V$. I personally don't think this is possible, induction doesn't work for formulas having $V$ in them, the proof that all naturals are in $V$ is not by induction. The usual proof is by defining a natural as an element of every set that has the empty set as an element that is closed under adjunction, of course this would be a set because $V$ is such a set and so every natural must be an element of $V$. So one needs to be careful from freely applying induction, because its not guaranteed for every formula. | |
| Jun 12, 2019 at 3:29 | history | edited | Master | CC BY-SA 4.0 |
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| Jun 12, 2019 at 2:57 | history | edited | Master | CC BY-SA 4.0 |
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| Jun 11, 2019 at 16:25 | history | edited | Master | CC BY-SA 4.0 |
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| Jun 11, 2019 at 16:19 | history | edited | Master | CC BY-SA 4.0 |
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| Jun 11, 2019 at 15:49 | comment | added | Master | Is it more clear now? | |
| Jun 11, 2019 at 15:49 | history | edited | Master | CC BY-SA 4.0 |
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| Jun 11, 2019 at 15:33 | comment | added | Master | The standard proof of the Reflection Theorem uses the set of all elements of $C$ of the least rank (See Set Theory, Chapter 12). $\Delta_1$ formulas are absolute for transitive sets so $V_\alpha^V=V_\alpha$, and that lets verify Replacement for $\alpha\leftrightarrow V_\alpha$. | |
| Jun 11, 2019 at 10:55 | comment | added | Zuhair Al-Johar | also what's the relevance of mentioning $\Delta_1$ point?, the most important is absence of the $V$ symbol. I'm still not sure of the proof. But I'll have a closer look at it and examine it further! | |
| Jun 11, 2019 at 5:08 | comment | added | Zuhair Al-Johar | what's the purpose of defining the set $\{x\in C|\forall y\in C(rank(x)\leq rank(y))\}$? this is the subset of $C$ that is the set of all elements of $C$ of the least rank. What's the relevance of this set to the proof. | |
| Jun 11, 2019 at 3:42 | history | edited | Master | CC BY-SA 4.0 |
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| Jun 11, 2019 at 0:21 | comment | added | Master | It is $\forall y\in C$. Is the answer sufficient now? | |
| Jun 11, 2019 at 0:12 | history | edited | Master | CC BY-SA 4.0 |
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| Jun 10, 2019 at 22:24 | comment | added | Zuhair Al-Johar | you have a typo at $x \in C | \forall y \in z ...$ I think $z$ is a typo, it should be $x$. | |
| Jun 10, 2019 at 18:41 | comment | added | Master | You are correct. I have adjusted my answer. Thank you for pointing that out. | |
| Jun 10, 2019 at 18:40 | history | edited | Master | CC BY-SA 4.0 |
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| Jun 10, 2019 at 18:11 | comment | added | Zuhair Al-Johar | Yes, true, but that doesn't mean that you proved that $V\models ZFC$ which is what you wrote in your answer [ you wrote: therefore $V\models ZFC$]. You can prove matters piecemeal as you said and this would "interpret" ZFC but not prove it. The difference is subtle, but its there! You need to correct your answer. | |
| Jun 10, 2019 at 15:42 | comment | added | Master | Just because $A$ does not prove $V\vDash ZFC$, doesn't mean $A$ does not prove $V\vDash\phi$ for any specific instance of $ZFC\,\phi$. For any specific instance of $ZFC\,\phi$, $ZFC\vdash \phi$ and so $\phi^V$. | |
| Jun 10, 2019 at 14:18 | comment | added | Zuhair Al-Johar | It is a theorem of Ackermann's that $V$ is not a model of $ZFC$. The reason is because ZFC is equivalent to Ackermann's over set sentences of Ackermann's, now if Ackermann's prove that $V\models ZFC$ then there would be a set sentence theorem of Ackermann's that "there exists of a model of ZFC", which simply means that ZFC would prove its own consistency, which is false per Godel's incompleteness arguments. | |
| Jun 10, 2019 at 14:11 | comment | added | Zuhair Al-Johar | You CANNOT prove Replacement in this theory simply because the parent theory (i.e. Ackermann's set theory) cannot prove replacement! This theory is just a fragment of Ackermann's set theory, so it cannot prove Replacement. However we may be able to "interpret" replacement. Notice that Ackermann's set theory cannot prove that $V$ is a model of ZFC, so this theory also cannot prove that. What can be done really is to check if Reinhardt proof did use class comprehension axioms in an essential way, in order to check if class comprehension is redundant or not. Cheers | |
| Jun 9, 2019 at 22:12 | comment | added | Master | I have added some more details. | |
| Jun 9, 2019 at 22:11 | history | edited | Master | CC BY-SA 4.0 |
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| Jun 9, 2019 at 18:54 | comment | added | Zuhair Al-Johar | generally speaking you are on the right track now. But I generally feel that proving this won't be easy, I think that what Reinhardt's proof is about. | |
| Jun 9, 2019 at 18:40 | comment | added | Master | You are correct. However, wouldn't $V$ at least satisfy Replacement for $\alpha\mapsto V_\alpha$, as $V$ is transitive and $x= V_\alpha$ is $\Pi_1$. Then it satisfies the reflection theorem and so we can find $\phi(x,y)^V\leftrightarrow \phi(x,y)^{V_\alpha}$, and so we can use $V_\alpha$ as a parameter? | |
| Jun 9, 2019 at 18:33 | comment | added | Zuhair Al-Johar | Yes, I was once thinking like that, but its not true, in order for $V$ to be a model of ZFC, you need all quantifiers in that statement to be bounded by $V$, this includes the formula of replacement, its not enough to just require that the parameters in $V$, and that the matrix of the formula doesn't use $V$, this would not make $V$ be the domain of a model of ZFC, so it would NOT provide the needed interpretation. | |
| Jun 9, 2019 at 18:29 | comment | added | Zuhair Al-Johar | the matter is a little bit subtle, I agree that it gives the appearance of having a proof by reflection, the problem is that barring $V$ from occurrence in the formulas of reflection will bar us from proving replacement for bounded by $V$ formulas, and to prove that $V$ is a model of ZFC, then all quantifiers in the statement of replacement must be really bounded by $V$. In order to achieve this you need some kind of definable sets in $V$ that do not use $V$, that bound the quantifiers and have the same effect of bounding by $V$. | |
| Jun 9, 2019 at 18:28 | comment | added | Master | The kind of formulas we want to use do not have the symbol $V$. If we want to prove $V\vDash ZFC$, we show that it satisfies Replacement for every formula with parameters in $V$. Those are exactly the kind of formulas Ackermann set theory allows us to use. | |
| Jun 9, 2019 at 18:22 | comment | added | Zuhair Al-Johar | If you show that for every $X \in V$, any definable function $F:X \to V$ implies that $F(X) \in V$, then yes this would definitely interpret Replacement over $V$! BUT the problem is how to prove that? Ackermann minus comprehension can only prove that statement for the kind of functions that are definable after formulae that do not have $V$ occurring in them and only when all their parameters are elements of $V$ also. It cannot show it for all kinds of formulas, in particular it cannot show it for bounded by $V$ formulas, which are the wanted formulas for a proof of replacement! | |
| Jun 9, 2019 at 16:05 | comment | added | Master | What we are trying to prove is first that $V\vDash ZFC$. To prove this we have to show that every definable function $F: V\rightarrow V$ has its image in $V$. By the definition of a function, its image is a subset of $V$ (You can't map using replacement, say $n\mapsto \kappa_n$, where $\kappa_n$ is the $n$th worldly cardinal, and therefore prove $Con(ZFC)$). | |
| Jun 9, 2019 at 11:01 | comment | added | Zuhair Al-Johar | what do you mean by $F(X) \subseteq V$ (by definition), you have an arbitrary class $X$ and $F(X) $is the image of $X$ under function $F$, why you think that images must be subsets of $V$? for example let $X$ be $\{V\}$ and let $F$ be the equality function, clearly $F(X)=X=\{V\}$ which is not a subset of $V$ and it is not an element of $V$. | |
| Jun 9, 2019 at 4:31 | comment | added | Master | I think I made a typo on the second to last line. Is it clearer now? | |
| Jun 9, 2019 at 4:31 | history | edited | Master | CC BY-SA 4.0 |
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| Jun 9, 2019 at 4:26 | comment | added | Zuhair Al-Johar | I didn't get the argument for Replacement. for example your $F(X)$ seems to be a subclass of the domain of $F$ it consists of elements of $dom(F)$ that have their images among their elements, I don't see why this $F(X)$ class should be a subset of $V$ for an arbitrary function $F$? Its not that clear. | |
| Jun 9, 2019 at 1:41 | history | edited | Master | CC BY-SA 4.0 |
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| Jun 8, 2019 at 22:20 | history | edited | Master | CC BY-SA 4.0 |
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| Jun 8, 2019 at 21:43 | history | answered | Master | CC BY-SA 4.0 |