Timeline for What is against having distinct membership relations on sets in the Platonic realm?
Current License: CC BY-SA 3.0
19 events
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| Mar 12, 2018 at 11:37 | comment | added | Joel David Hamkins | I view the argument I gave as a version of Tony Martin's categoricity argument in philpapers.org/rec/MARMUO. Basically, I view this result as a ZFC/GBC version of Martin's philosophical argument. | |
| Mar 12, 2018 at 11:31 | comment | added | Joel David Hamkins | @MonroeEskew I posted this answer because I find the result interesting and related to the ideas expressed by the OP. But also (1) the case without a combined language is basically trivial, since one can clearly combine two models on a common domain; (2) if one wants to understand $\in$ and $\in^*$ as providing a mathematical foundation, then they should provide a foundation for each other, and therefore satisfy set theory in the combined language. For example, the result applies when $\in^*$ is definable in $\in$ and vice versa, and this is interesting. | |
| Mar 12, 2018 at 11:24 | comment | added | Monroe Eskew | The OP said "two theories each referring to one of these membership relations," and did not speak of a combined language. | |
| Mar 11, 2018 at 18:56 | vote | accept | Zuhair Al-Johar | ||
| Mar 11, 2018 at 18:56 | |||||
| Mar 10, 2018 at 18:20 | comment | added | Zuhair Al-Johar | this means that the analogy with arithmetical operators that I've made for motivating having multiple membership relations on sets would break since the arithmetical operators are indeed used in combined language theories. | |
| Mar 10, 2018 at 18:00 | comment | added | Joel David Hamkins | If the relations are not available as classes to one another, then the situation is basically trivial as Will Sawin pointed out above. Any two countable models of set theory can be placed as relations on the same underlying countable set. | |
| Mar 10, 2018 at 17:59 | comment | added | Joel David Hamkins | @Zuhair Yes, I was using the combined language and the map $\pi$ at each step, taking the $\in$-power set on the left and the $\in^*$-power set on the right. These must be isomorphic, since you can pull any subset over to the other side, using that the relations are each available as classes to the other relation. | |
| Mar 10, 2018 at 15:46 | comment | added | Zuhair Al-Johar | @NotMike ...[cont.] which is a combined instance. Of course, if both relations can see each other (which is what using a combined language [and axiomatization] would entail) then we cannot have these two relations, my idea is to have each membership relation obeying $\text{ZF}$ axioms without seeing the other. So my approach is uncombined. I think this is possible. | |
| Mar 10, 2018 at 15:42 | comment | added | Zuhair Al-Johar | NotMike I don't see enough details in Joel's answer, lets map the $\in$ empty set to the $\in^*$ empty set this is easy since we have pairing, we can easily build the Kuratowski ordered pair of those, but how can I go up with powers, to do that I imagine I need. $\pi(x)=k \leftrightarrow \forall y(y \in^* k \leftrightarrow \exists m \in x(y=\pi(m)))$, but how we prove the existence of such a set, which of the [uncombined] ZF axioms of $\in^*$ or $\in$ is used to define that mapping for each x? Unless he is taking the $\in^*$ power of $\pi(x)$ and then separating using a formula that uses $\in$ | |
| Mar 10, 2018 at 11:23 | comment | added | Not Mike | @Zuhair it's constructed by induction. Start by observing that the empty-set is a distinguished object of $ZF$. So begin with $\pi$ mapping the first empty-set to the other. From there use the prescription given in the answer to handle successor and limit stages. | |
| Mar 10, 2018 at 10:03 | comment | added | Zuhair Al-Johar | This question is to Joel, I don't see how can I define an isomorphism $\pi$ between the stages $V_{\alpha}$ and $V_{\alpha^*}$ I can have an object in $P^{sets}$ that is $\in $ empty but not $\in^*$ empty. | |
| Mar 10, 2018 at 9:40 | comment | added | Zuhair Al-Johar | @WillSawin I didn't understand what you are trying to say, what do you mean by "of course not", not what? and what is the relation of not having a relation between $\in_1$ and $\in_2$ to that conclusion of yours? | |
| Mar 10, 2018 at 8:18 | comment | added | Will Sawin | @Zuhair With no relation between $\in_1$ and $\in_2$, of course not. Simply take two countable models of ZF and fix a bijection between them. | |
| Mar 10, 2018 at 7:21 | comment | added | Zuhair Al-Johar | What if the language is not combined? Can we have two theories $\text{ZF}_1, \text{ZF}_2$ written in $L_{\langle V, \in_1 \rangle}, L_{\langle V, \in_2 \rangle}$ both (as named) obeying all axioms of $\text{ZF}$ over the same domain $V$ and yet Choice holds in one and fails in the other? Notice that $P^{\in_1}$ and $P^{\in_2}$ are not collections of sets, i.e. the primitive ordered pairs are not sets, so how can you get rules about them even in a combined theory? | |
| Mar 10, 2018 at 3:06 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Mar 10, 2018 at 2:47 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Mar 10, 2018 at 0:38 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Mar 10, 2018 at 0:19 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Mar 10, 2018 at 0:09 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |