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Given a model of set theory $V$ there are various ways to construct a model in which the Axiom of Choice holds, such as Gödel's constructible universe $L^V$ or by using forcing*. I'm wondering if any of these constructions have a nice universal property in the sense of category theory.

In particular are there any adjoints to the inclusion 2-functor $\mathcal{ZFC}\to\mathcal{ZF}$?

(Where $\mathcal{ZF}$ is the 2-category whose objects are either

  1. Models of ZF
  2. Toposes

and whose morphisms are either

  1. Geometric morphisms
  2. Logical functors
  3. Elementary embeddings**

and where $\mathcal{ZFC}$ is the full subcategory on the objects that obey Choice.)

I think that the concept of the "minimal model" might allow one to construct a right adjoint which is similar in character to the constructable universe. Forcing goes "the other way" by adding sets rather than restricting them, so I suspect it might give a left adjoint.

*Or not? See comments.

**Asaf Karagila notes that there can't exist an elementary embedding between a model where Choice holds and one where it doesn't. So there can't be an adjunction in this case, because its unit or counit would sometimes have to be such a morphism. But perhaps there's some other kind of morphism between models of ZF that does allow an adjunction?

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    $\begingroup$ @MikeShulman One can force any given set to be well-orderable (or even countable, an extreme case of this), and this means that in many cases, one can force AC. For example, over any model of the form $L(\mathbb{R})$, one can force AC, essentially by adding a generic well-ordering of the reals. If DC holds, one can do this without adding reals. Meanwhile, there are models of ZF, such as the Gitik model where every $\aleph_\alpha$ is singular, where ZFC fails in every extension with the same ordinals, and so in such a case, there can be no forcing extension with ZFC. $\endgroup$ Mar 22, 2018 at 16:38
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    $\begingroup$ I wonder whether the question of the OP can be answered by means of the fact that ZF and ZFC are not strictly bi-interpretable. Indeed, one can prove that different set theories extending ZF are never bi-interpretable. I wonder what is the topos-theoretic version of this result? $\endgroup$ Mar 22, 2018 at 16:40
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    $\begingroup$ @Mike: Blass in his paper "injectivity projectivity and the axiom of choice" defined an axiom SVC, small violations of choice, which is equivalent to the statement "AC can be forced". This holds in all symmetric models, or models of the form L(X) or HOD(X) or generally V(X) when V a model of ZFC. But it does not hold in general for models of ZF, of course. Be it via class symmetric extensions of certain kinds, or just by assuming certain axioms hold and/or fail (e.g., in my thesis I show that the failure of KWP is consistent, and it implies the failure of SVC). $\endgroup$
    – Asaf Karagila
    Mar 22, 2018 at 19:30
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    $\begingroup$ @Mike: Well, I wouldn't say it this way. Rather if choice fails because of one set, we can force it back. But if choice fails because of a proper class of sets, then it's impossible to fix. $\endgroup$
    – Asaf Karagila
    Mar 23, 2018 at 8:32
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    $\begingroup$ Well, the reason I ask is that an elementary embedding does not exist between models with different theories. So no model of ZFC embeds elementary into a model of non-AC, and vice versa. So your inclusion functor would be sort of degenerate. $\endgroup$
    – Asaf Karagila
    Mar 25, 2018 at 13:10

2 Answers 2

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Too long to be a comment but not really answer either. This is an interesting question because it is phrased in terms of category theory. There are various way of forcing the full axiom of choice over models of $ZF$ which already satisfy weak forms of choice.

For example starting with $L(\mathbb{R})$ and assuming $AD+DC$, one may force the full axiom of choice using $\mathbb{P}_{max}$ forcing. This forcing will add a wellordering of the reals of length $\aleph_2$ and thus achieve $\Theta=\aleph_3$ ($\Theta$ is the sup of the length of the prewellordering of the reals, useful in non-$AC$ context). The wellordering of the real is not added in the usual way, using say $Coll(\omega_1,<\mathbb{R})$. The reason behind forcing $AC$ here traces back to the existence of stationary-co-stationary subsets of $\omega_1$. In fact, if there exists a stationary-co-stationary subset of $\omega_1$ then there is an $\omega_1$-sequence of distinct reals (this is already a manifestation of a weak form of choice if the elements of the model are all ordinal definable from reals)

The above technique is not general enough because forcing choice will disturb the cardinal arithmetic of the model in general. We believe we have proved recently that if one starts with $AD+DC$ in $L(\mathbb{R})$ then one may force $AC$ while making the continuum $\aleph_3$. This is very different from $\mathbb{P}_{max}$ forcing but it is still entirely possible that there is some deep structure which accounts for very general methods on how to force choice, under large cardinal hypothesis. Forcing various degrees of generalizations of $DC$ is crucial to achieve the result.

I would be very interested to see how this translates into a category theoretic framework and see if there is a way to gauge the "universality" of the method.

As mentionned by Joel, starting from $ZF+DC$ one may force $AC$ using $Coll(\omega,<\mathbb{R}).$ By forcing theory this wellorders the reals, does not add reals and one obtains $CH$ from it.

Finally, in a different direction, David Pincus has shown how to transform a model of a statement $\phi$ into model of $\phi+DC$, see the following article: Adding Dependent Choice. The contents of the article seem to be more amenable to a category theoretic approach than the results we've mentioned above.

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  • $\begingroup$ What is $\text{Coll}(\omega,\mathord{<}\mathbb{R})$? Does it mean the same thing as $\text{Coll}(\omega_1,\mathbb{R})$? $\endgroup$ Mar 23, 2018 at 18:15
  • $\begingroup$ Yes, I meant to write $Coll(\omega_1,\mathbb{R})$, which under $DC$ wellorders the continuum in length $\aleph_1$. $\endgroup$ Mar 23, 2018 at 18:29
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    $\begingroup$ Uhh, are you sure about that stationary-co-stationary? Because if you just look at Solovay's model obtained by the least inaccessible $\kappa$ being collapsed to be the new $\omega_1$, then you did not kill the stationarity of subsets of $\kappa$ in $L(\Bbb R)$, but there is no $\omega_1$ sequence of reals. $\endgroup$
    – Asaf Karagila
    Mar 23, 2018 at 19:31
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    $\begingroup$ Also Pincus' paper is about relatively specific models. So I don't agree that it is more amenable than claims about $L(\Bbb R)$. $\endgroup$
    – Asaf Karagila
    Mar 23, 2018 at 19:34
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Following up on my remarks in the comments, allow me to answer from the perspective of model-theoretic interpretations of theories. I view interpretations of theories as providing particularly strong forms of the desired functors between these categories.

Specifically, an interpretation of one theory $S$ in another theory $T$, is a uniform way of defining a model of $S$ inside any model of $T$. Given a model of the latter theory $M\models T$, one defines a domain $N$ and functions and relations on this domain, such that with this structure it becomes a model of the first theory $N\models S$.

Two theories are mutually interpretable, if each of them is interpreted in the other. So in any model of the one theory $M\models T$ you can define model of the other theory $N\models S$, and inside $N$ you can define a model $\bar M\models T$ of the first theory again.

The thing to notice is that with mutual interpretability, there is no insistence that these interpretations are inverse of each other, and it could be that $M$ and $\bar M$ are not much related. Perhaps this makes them rather un-adjoint-like.

A much stronger notion, therefore, imposes the uniform inverse requirement. Specifically, theories $S$ and $T$ are bi-interpretable if they are mutually interpretable in such a way that the interpreted model $\bar M$ arising in $N$ is isomorphic to $M$ and furthermore, isomorphic by an isomorphism that is definable in $M$, and vice versa in the other direction.

The relevance of this for your question is that ZF and ZFC are not bi-interpretable.

Theorem. Distinct extensions of ZF are never bi-interpretable.

Thus, one cannot transform ZF models and ZFC models into one another in such a way that they form a bi-interpretation, and I take this to be a kind of negative answer to a strong version of the question.

I recently made a blog post providing a proof and further discussion of this theorem and related matters:

Different set theories are never bi-interpretable.

(The theorem follows from results of Albert Visser in his 2006 paper, "Categories of theories and interpretations." In addition, there is a nice automorphism group model-theoretic argument of Ali Enayat showing for the specific case of ZF and ZFC that they are not bi-interpretable. Follow the link at my blog.)

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  • $\begingroup$ Luckily I don't think we need a bi-interpretation. We're not looking for an adjunction between models of the set theory; we're looking for an adjunction between the categories of the models. This adjunction will induce a map from each model of ZF to the corresponding model of ZFC (or in the other direction depending on which way around the adjunction is). So we only need a morphism going one way rather than both. The category of models and interpretations looks like an interesting place to look for the adjunction in. $\endgroup$ Mar 28, 2018 at 8:01
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    $\begingroup$ I think I understood that. My point was that if you take models of set theory seriously as foundational realms, then you'd want the image model of a model to be accessible to the original model, in the way that the bi-interpretation insists upon. Otherwise the applications of your adjunction live in a place where no model can see it. $\endgroup$ Mar 28, 2018 at 10:21
  • $\begingroup$ Ignore my above comment, I hadn't drunk enough coffee yet (I was thinking that bi-interpretability was a relationship between models, whereas in fact it's a relationship between theories). Am I right in thinking that ZF has an obvious interpretation in ZFC (just take the whole model)? So what we're looking for is an interpretation of ZFC in ZF with some nice properties. In this case (where one interpretation is trivial) the definition of bi-interpretation says that every model of ZF should be isomorphic to the model of ZFC it induces. Clearly this is stronger than what we want. $\endgroup$ Mar 28, 2018 at 13:07
  • $\begingroup$ But it would be nice if this was true whenever out original model obeyed choice. Do you know of an interpretation of ZFC in ZF that does this? i.e. Is there an interpretation of ZFC in ZF such that whenever you apply it to a model of ZF that happens to obey Choice the induced model is definably isomorphic to the original? That question is probably what I would have asked if I was trained to think in terms of logic rather than category theory. $\endgroup$ Mar 28, 2018 at 13:16
  • $\begingroup$ Yes, there is such an interpretation: if AC holds, take all of V, otherwise go to L. $\endgroup$ Mar 28, 2018 at 21:26

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