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I am curious whether or not the following axiom is independent of Hamkins's axioms for the Set-Theoretic Multiverse. Hamkins's axioms can be found here on pages 1-2 and here on pages 24-26.

Consider the following axiom: If $(M,\in_M)$ and $(N,\in_N)$ are set-theoretic universes, then there exists a universe $(W,\in_W)$ such that there exists injective functions $j:M\to W$ and $i:N\to W$ where $(W|_{j(M)}, \in_W) \cong (M,\in_M)$ and $(W|_{i(N)},\in_W) \cong (N,\in_N)$.

Remark: Independence might not be the correct work I am looking for (since the Set-Theoretic Multiverse is a foundational structure). I am looking to see whether this axiom (as well as its negation) is coherent in the same sense that the Multiverse axioms are coherent (i.e. in the same way that the collection of countable computably-saturated models of ZFC satisfies the multiverse axioms).

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    $\begingroup$ There are several of multiverse notions; Hamkins, Steel, Woodin and Väänänen. Each of them has his own definition of a multiverse (granted, Steel and Woodin were mainly interested in the generic multiverse, but still). It's a real multiverse of multiverses. $\endgroup$ – Asaf Karagila Dec 29 '14 at 22:25
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    $\begingroup$ Curious bystander: How would this be an analogue of Vopenka? I know there are many statements of Vopenka's principle, but the one I'm most familiar with says that if you have a proper class of models, then there's a morphism from one of them to another. I don't see an analogue of the proper class of models here... $\endgroup$ – Tim Campion Dec 29 '14 at 22:28
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    $\begingroup$ @TimCampion I think the multiverse is meant to be the analogue of "proper class of models," and interpretations are the morphisms. $\endgroup$ – Noah Schweber Dec 29 '14 at 22:30
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    $\begingroup$ @KyleGannon, by "coherent" do you mean "consistent?" I thought that at first, but now I'm not quite sure. (Also, I misunderstood you; I've deleted a now-irrelevant comment.) $\endgroup$ – Noah Schweber Dec 29 '14 at 22:31
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    $\begingroup$ The axiom really looks like a variant of the joint embedding property, not of Vopěnka’s principle. $\endgroup$ – Emil Jeřábek Dec 29 '14 at 22:34
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Some time ago, I looked into arithmetical consequences of the set-theoretic multiverse.

I started with the assumption that there are numbers that are standard, which I will denote $\omega$ and that we have some useful conception of subsets of $\omega$, I found that if two multiverses are compatible in the sense that they are both elements of some larger universe then they must have the same standard system. There is also an interesting converse to this that I discuss in my post.

In detail, if $U$ is a universe then elements $n$ of $\omega^U$ code certain subsets of $\omega$. Namely, $n$ codes the set $x_n$ of all standard $i \in \omega$ such that the $i$-th binary digit of $n$ is $1$. The collection of all sets coded by elements of $\omega^U$ in this way is called the standard system of $U$ and is denoted $\operatorname{SSy}(U)$.

If a universe $U$ is an element of a larger universe $V$ then we must have $\operatorname{SSy}(U) = \operatorname{SSy}(V)$. Indeed, because $V$ can recognize that $\omega^U$ is a model of a reasonable arithmetic theory, $V$ realizes that $\omega^V$ is an initial segment of $\omega^U$ since $V$ believes that $\omega^V$ is the standard model of arithmetic. From this, we find that $U$ and $V$ have the same standard system. Indeed, if $p$ is any nonstandard element of $V$ then every $x \in \operatorname{SSy}(V)$ is coded by some $n < 2^p$, and such $n$'s are both in $U$ and $V$.

Note that the assumptions I used are debatable, as discussed by Joel in this post. I never got around to writing the sequel of this post where I weaken the hypotheses somewhat. Perhaps I will find some time now that I am reminded of this...

Also note that compatibility, as defined above, is much stronger than your joint embedding property. However, the underlying reason why I chose to look at compatibility rather than embeddings remains: how would we know that such embeddings exist unless such embeddings exist in some larger universe?

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Let me add to Noah's remarks that your axiom (and Noah's) is true in the model that Victoria Gitman and I provide in our paper A natural model of the multiverse axioms, which you also had mentioned. In that paper, we prove that the collection of all countable computably saturated models of ZFC satisfies all the other multiverse axioms. It also satisfies your joint embedding property axiom, because of the main theorem of my article, Every countable model of set theory embeds into its own constructible universe, which shows that every countable computably saturated model of ZFC is universal for all countable acyclic binary relations. So in fact all these models are isomorphic to submodels of each other.

Theorem. In any multiverse consisting of countable models, satisfying the multiverse axioms, the joint embedding property also holds.

Proof: It follows from the well-founded mirage axiom that the multiverse must have an ill-founded model $W$, and the ordinals of $W$ must therefore contain a copy of $\mathbb{Q}$. It follows by my embeddings theorem that $\langle W,\in^W\rangle$ is universal for all countable acyclic binary relation. In particular, any two countable models of set theory embed into $W$, and so we have the joint embedding property. QED

In fact, it follows from the well-founded mirage axiom that every model $W$ of the multiverse is ill-founded, and hence universal for all countable models of set theory.

But let me also point out that it follows from results in that paper that this notion of embedding is probably not what you should mean. A mere $\in$-homomorphism need not map ordinals to ordinals, and can map the empty set to a nonempty set and so on. Probably one wants to consider at least $\Delta_0$-elementary embeddings, if not much more than this.

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  • $\begingroup$ Thanks for your response. Might you know of a construction where these facts fail? $\endgroup$ – Kyle Dec 29 '14 at 23:06
  • $\begingroup$ Gitman and Fuchs and I are currently preparing a paper on $\omega_1$-like models of set theory for which the embedding comparability theorems fail, and it is conceivable to me that one could build a multiverse using such models in which your joint embedding property fails. $\endgroup$ – Joel David Hamkins Dec 29 '14 at 23:13
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The answer seems to be "no": since none of the multiverse axioms involve two universes at once, the union of two multiverses is a multiverse. This lets you easily (EDIT: As Joel points out, this is totally false: it is not at all easy, although it may be true) construct counterexamples to the claim.

Meanwhile, if we add the (in my opinion, desirable) axiom that for every pair of universes $M$, $N$ there is a $W$ in which both $M$ and $N$ are countable sets, then the answer is clearly "yes;" and moreover, these two statements are equivalent (given $W$ interpreting $M$ and $N$, just consider a $V$ in which $W$ is countable.

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  • $\begingroup$ Although of course in the second paragraph, I'm assuming that by "interpretable" you mean "interpretable with parameters." If you mean parameter-free interpretability, then I'm pretty sure the answer goes back to "no." $\endgroup$ – Noah Schweber Dec 29 '14 at 22:29
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    $\begingroup$ I don't understand the claim of your first paragraph. If one takes the union of two multiverses of countable models, it will still satisfy the desired joint embedding property, because of my theorem on the fact that countable models are always comparable by embeddability (the multiverses must have models with nonstandard ordinals, and hence there will be universal objects there). Could you explain what you had in mind? $\endgroup$ – Joel David Hamkins Dec 29 '14 at 23:02
  • $\begingroup$ I was visualizing a pair of multiverses all of whose elements were uncountable universes (with illfounded $\omega$); I'm pretty sure that we can get a failure of joint embeddability with such a pair. Is this not the case? $\endgroup$ – Noah Schweber Dec 29 '14 at 23:53
  • $\begingroup$ Probably, and I expect so, but I don't see it as immediate. After all, it was some work even to get uncountable models that were incomparable by embeddability at all. $\endgroup$ – Joel David Hamkins Dec 29 '14 at 23:56

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