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This question is in connection with the question that I've asked at:

Where do models of false theories exist?

The answer to that question was that any consistent theory can have its primitives be re-interpreted in such a manner as to come true. So the difference between a false theory and a true theory is one of reference, a true theory is one whose sentences are satisfiable in the part of the Platonic realm that it refers to.

Now according to that answer, I'll pose the following possibility and the question is what is against that possibility:

Now let's assume that there exists a Platonic world $P^{sets}$ of all sets, and two Platonic worlds $P^{\in_1}$ , $P^{\in_2}$ of primitive ordered pairs of sets, these are taken to represent distinct membership relations between sets. So the ordered pairs in realms $P^{\in_1}$, $P^{\in_2}$ only have sets as their projections, so $P^{sets}$ is their domain, so they represent two kinds of membership relations between sets $\in_1$ and $\in_2$ relations defined on the same domain, as:

$y \in_1 x \iff \exists p \in^* P^{\in_1} (p=\langle y,x \rangle)$

$y \in_2 x \iff \exists p \in^* P^{\in_2} (p=\langle y,x \rangle)$

So for example the sentence $\exists x \forall y (y \not \in x)$ would be:

$\exists x \in^* P^{sets} \forall y \in^* P^{sets} (\not \exists p \in^* P^{\in_1} (p=\langle y,x \rangle))$

$\exists x \in^* P^{sets} \forall y \in^* P^{sets} (\not \exists p \in^* P^{\in_2} (p=\langle y,x \rangle))$

Where $\in^*$ is the membership relation between sets and $P^{sets}$ and between ordered pairs of sets and the realms $P^{\in_1}, P^{\in_2}$.

Now since we are having two membership relations on sets defined after two Platonic realms, then we can have two theories each referring to one of these membership relations, so no confusion of reference would raise (as far as each theory is speaking correctly about the part of the Platonic realm that it refers to), and so both theories would be TRUE in that Platonic world. Accordingly we can have both membership relations obeying all rules of $\text{ZF}$ and yet one of them obeying $\text{Choice}$ while the other negating it.

So this would mean that the answer to as whether choice or negation of choice is true about membership in sets, is to say that there are two kinds of membership in sets, one fulfills choice and the other negates it.

I don't see anything in the definition of a true theory [from a Platonic perspective] that can go against that possibility. Why should there be just one kind of membership in sets? there is no rule to the effect that no two distinct relations in the Platonic world can have the same domain, actually, this is not the case with the standard model of arithmetic, for it does have distinct relations having the same arity defined on the same domain of standard naturals (exp: the binary relations $Successor$ ,$<$; the ternary relations $+ ,\times $) so why not have the same situation with sets?

Along the same lines of this argument, we may have two membership relations one obeying $\text{CH}$ and the other negating it, on the SAME domain of all sets.

This question is intended to be answered from a Platonistic perspective.

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    $\begingroup$ It would seem to me that only a young and naive Platonist would be willing to argue for the existence of an idealized true theory. Since each abstract object which interprets a base theory can be considered an idealized object representing the entirety of the statements it satisfies. The very fact that you would have differing interpretations of a weak base theory which disagree on certain statements would seem to me to imply they are approximations to distinct and incomparable abstract objects. $\endgroup$ – Not Mike Mar 9 '18 at 22:59
  • $\begingroup$ There is no such thing as a Platonic realm. $\endgroup$ – Qfwfq Mar 9 '18 at 23:12
  • $\begingroup$ @NotMike I'm speaking about membership relations of $\text{ZF}$ and I don't think $\text{ZF}$ is a weak base theory? $\endgroup$ – Zuhair Al-Johar Mar 10 '18 at 10:08
  • $\begingroup$ @Zuhair No, you're speaking about collections of ordered pairs, whose nature you presuppose as being paradoxical relative to an inconsistent interpretation of mathematical platonism. $\endgroup$ – Not Mike Mar 10 '18 at 10:58
  • $\begingroup$ @Zuhair and form the perspective of many set-theorists, $ZF$ is a weak collection of axioms. $\endgroup$ – Not Mike Mar 10 '18 at 11:07
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Putting on my Platonist hat for a while (and it's a very comfortable hat), I'd answer the question as follows: There's nothing wrong with having and studying two or more relations, like $\in_1$ and $\in_2$ in the question. But I wouldn't want to call both of them "membership"; like many people, I get confused when the same word is used for two different things. I'd reserve the name "membership" for actual membership, which seems to be what you called $\in^*$ in the question.

If I take off my Platonist hat and put on my formalist hat or my materialist hat (hats which I don't wear often because they give me headaches), then I'd answer as in Qfwfq's comment.

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  • $\begingroup$ $\in_1$ is named as: "set membership of the first kind", $\in_2$ is named as: "set membership of the second kind", and $\in^*$ is "realm membership", and realms are not sets. So they are indeed named differently, but they do share a common word (namely membership) because they do possess major overlaps (except for $\in^*$ which is in some sense different, I usually think of it as a mereologically based relation between atoms and totalities of atoms). I think it is plausible to name distinct highly overlapping entities by overlapping distinct names so that names copy the nature of the named. $\endgroup$ – Zuhair Al-Johar Mar 10 '18 at 9:58
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It turns out that one cannot have two fundamentally different parallel set membership relations $\in$ and $\in^*$, if they both satisfy ZF with respect to the common language, for in this case they must in fact be isomorphic. In particular, in this situation, they cannot satisfy different theories.

Theorem. If $\in$ and $\in^*$ are membership relations each satisfying the ZF axioms in the combined language, then $\langle V,\in\rangle$ is isomorphic to $\langle V,\in^*\rangle$.

Proof. One can see that $\in^*$ must be well-founded, since for every set $x$ the $\in^*$-class $\{y\mid y\in x\}$ has an $\in^*$-minimal element, by the $\in^*$-foundation axiom, and such an element is an $\in^*$-minimal element of $x$. Similarly, $\in$ is well-founded with respect to $\in^*$.

If one also knows that the relations $\in^*$ and $\in$ are set-like with respect to each other, then we may consider the Mostowski collapse $$\pi(x)=\{\pi(y)\mid y\in^* x\},$$ which is an isomorphism of $\langle V,\in^*\rangle$ with some transitive class $\langle M,\in\rangle$. But in fact, we must have $M=V$ since if every element of a set $z$ is $\pi(y)$ for some $y$, then let $x$ be the set with $\in^*$-elements $y$ whenever $\pi(y)\in z$. This set exists by replacement, using that $\in$ is set-like with respect to $\in^*$, and it follows that $\pi(x)=z$. So by $\in$-induction, every set is in $M$, and we have the desired isomorphism.

But for the general case, one doesn't know at first that the relations are set-like and so more care is needed. Consider the situation of $$\pi:\langle V_\alpha,\in\rangle\cong\langle V^*_{\alpha^*},\in^*\rangle,$$ where $\alpha$ is an $\in$-ordinal and $\alpha^*$ is an $\in^*$-ordinal and $V_\alpha$ and $V^*_{\alpha^*}$ are the rank-initial segments of the universe as constructed with respect to the two membership relations. Since transitive sets are rigid, these isomorphisms are unique when they exist. We can always extend to one more level by considering the power sets, which map across by their pointwise action. And we can take unions at limit stages. It cannot be that one side runs out of ordinals before the other, for in this case we would have a bijection of the whole universe to a set, either in the $\in$-universe or in the $\in^*$ universe. So this provides the desired isomorphism, as well as a proof that the relations are in fact set-like. $\Box$

Since isomorphic models have the same theory, it follows that:

Corollary. If $\in$ and $\in^*$ are membership relations satisfying the ZF axioms in the combined language, then $\langle V,\in\rangle$ has the same theory as $\langle V,\in^*\rangle$.

I think this theorem is classically known. Similar ideas are used in my paper:

Hamkins, Joel David; Kikuchi, Makoto, Set-theoretic mereology, Log. Log. Philos. 25, No. 3, 285-308 (2016). DOI:10.12775/LLP.2016.007, ZBL1369.03047.

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  • $\begingroup$ 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? $\endgroup$ – Zuhair Al-Johar Mar 10 '18 at 7:21
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    $\begingroup$ @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. $\endgroup$ – Will Sawin Mar 10 '18 at 8:18
  • $\begingroup$ @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? $\endgroup$ – Zuhair Al-Johar Mar 10 '18 at 9:40
  • $\begingroup$ 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. $\endgroup$ – Zuhair Al-Johar Mar 10 '18 at 10:03
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    $\begingroup$ @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. $\endgroup$ – Not Mike Mar 10 '18 at 11:23

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