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Given two logical systems, $L_1$ and $L_2$, in which every formula of $L_1$ is also a formula of $L_2$, $L_2$ is said to be a conservative extension of $L_1$ iff the set of theorems of $L_1$ are precisely the set of theorems of $L_2$ that are formulas of $L_1$.

Denote by $\text{ZFC}_{\sigma}$ the ZFC set theory enriched with a $1$-place function symbol '$\sigma$', called the global selector, and extended with the following axiom, called the axiom of global choice for ZFC.

For every non-empty set $z$, $\sigma(z)$ is a member of $z$.

According to [Fraenkel] (p. 73), $\text{ZFC}_{\sigma}$ is a conservative extension of ZFC.

Denote by NBG+ the NBG set theory extended with the following axiom, called the axiom of global choice for NBG.

There exists a function $\sigma$ whose domain contains all non-empty sets, and such that for every non-empty set $z$, $\sigma(z) \in z$.

(Note that the language of NBG+ is not enriched with a symbol $\sigma$.)

According to [Fraenkel] (p. 134), NBG+ is a conservative extension of ZFC.

Denote by $\text{NBG}_{\sigma}$ the NBG set theory enriched with a $1$-place function symbol '$\sigma$', called the global selector, and extended with the following axiom.

For every non-empty set $z$, $\sigma(z)$ is a member of $z$.

Questions

  1. Is $\text{NBG}_{\sigma}$ a conservative extension of NBG+?
  2. Is $\text{NBG}_{\sigma}$ a conservative extension of NBG?
  3. Is NBG+ a conservative extension of NBG?

Bibliography

[Fraenkel] Fraenkel, A. A., Bar-Hillel, Y., Levy, A. Foundations of Set Theory. 2nd Revised Edition. Elsevier. 1973

Evan Aad
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  • What's the relationship between $\mathsf{ZFC}_\sigma$ and $\mathsf{ZFC}+V=\mathsf{HOD}$? – Alessandro Codenotti Jun 26 '19 at 20:35
  • @AlessandroCodenotti: What's V? What's HOD? And what do they have to do with my questions? – Evan Aad Jun 26 '19 at 20:37
  • $V=\mathsf{HOD}$ is equivalent to "there exist a definable global well ordering of the universe $\varphi(x,y)$" (from which you get a global choice function simply by sending every nonempty set to its $\varphi$-minimal element). $V$ is the class of all sets, $\mathsf{HOD}$ is the class of hereditarily ordinal definable sets, whose definition won't fit in this comment – Alessandro Codenotti Jun 26 '19 at 20:42
  • @AlessandroCodenotti: I'm afraid I'm not a set theorist. These concepts mean little to me. I don't know how to answer the question whether V = HOD. Moreover, I don't see how the answer to this question, whether yes or no, will help me resolve my questions. – Evan Aad Jun 26 '19 at 20:47
  • How is $NBG_+$ different from $NBG_\sigma$? Aren't both given the first-order axiom $\forall z\ne\emptyset.\sigma(z)\in z$? – R. Burton Jun 26 '19 at 20:50
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    @AlessandroCodenotti: The formulas of NBG+ are precisely the formulas of NBG, whereas the formulas of $\text{NBG}_{\sigma}$ are a superset of the formulas of NBG, since they may contain the extra function symbol $\sigma$. – Evan Aad Jun 26 '19 at 20:54
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    In $ZFC_\sigma$ do you include instances of selection/ replacement that contain $\sigma$? – hmakholm left over Monica Jun 26 '19 at 22:00
  • @HenningMakholm: $\text{ZFC}_{\sigma}$ is precisely ZFC enriched with a distinguished $1$-place function symbol $\sigma$ and extended with the axiom of global choice. Nothing more, nothing less. – Evan Aad Jun 26 '19 at 22:02
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    @EvanAad: So one can't use the $\sigma$ symbol in the formula in selection and replacement, or can one? (It doesn't clarify much when you respond to a clarifying question by essentially repeating the same unclear description I'm asking you to clarify ...) – hmakholm left over Monica Jun 26 '19 at 22:25
  • @HenningMakholm: $\sigma$ is a fixed function, the same way $\emptyset$ is a fixed set. – Evan Aad Jun 26 '19 at 22:30
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    @EvanAad: Why are you not answering whether you can use that constant in instances of the replacement and selection axioms? How is it so hard just to answer yes or no? – hmakholm left over Monica Jun 26 '19 at 22:31
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    (And $\varnothing$ is not a symbol in the language of ZFC, which is a clear difference from when you say that $\sigma$ is a new symbol you are adding to the language). – hmakholm left over Monica Jun 26 '19 at 22:32
  • @HenningMakholm: This is an astute observation. Then let me rephrase. $\sigma$ is a fixed function the way $\in$ is a fixed relation. – Evan Aad Jun 26 '19 at 22:34
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    Is $\forall x\exists y\forall z(z\in y\leftrightarrow \exists w(w\in x\land z=\sigma(w)))$ -- or in other words "for all $x$ the set ${\sigma(w)\mid w\in x}$ exists" -- an axiom of your $ZFC_{\sigma}$? $$\huge\text{YES OR NO?}$$ – hmakholm left over Monica Jun 26 '19 at 22:39
  • @HenningMakholm: No, it is not an axiom. The only axiom that includes the $\sigma$ function is the axiom of global choice. – Evan Aad Jun 26 '19 at 22:45
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    Okay. Why did I have to waste half an hour getting you to answer that? – hmakholm left over Monica Jun 26 '19 at 22:45
  • @HenningMakholm: Actually, I should be more precise in what I meant by "the axiom of global choice", since I didn't use this title in the section where I described the system $\text{NBG}_{\sigma}$. What I meant to say was that the only axiom that includes the $\sigma$ symbol is the following: "For every non-empty set $z$, $\sigma(z)$ is a member of $z$." Equivalently, in symbols: $\forall z(z\neq\emptyset\implies\sigma z\in z)$. – Evan Aad Jun 26 '19 at 22:54
  • @HenningMakholm: Equivalently, without the $\emptyset$ notation: $\forall z\big(\exists x(x\in z)\implies\sigma z\in z\big)$ – Evan Aad Jun 26 '19 at 23:00
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    One point here is that it is almost trivially obvious that this addition of $\sigma$ produces a conservative extension when you're not adding new instances of selection/replacement. You can take any model of ZFC whatsoever and make it into a model of ZFC${}\sigma$ by defining an appropriate interpretation $\sigma$. This does not involve adding any new _sets or changing the relation between the sets you already have, so every sentence in the language of ZFC retains its truth value, and there is now only your one new axiom to satisfy. And it is just the same between NBG and NBG${}_\sigma$. – hmakholm left over Monica Jun 26 '19 at 23:10
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    In other words, you cannot use your new $\sigma$ symbol for anything meaningful in proofs as long as you don't allow it to appear inside the set builder notation. – hmakholm left over Monica Jun 26 '19 at 23:11
  • @HenningMakholm: ZFC is sound and complete, correct? Is NBG too sound and complete? – Evan Aad Jun 26 '19 at 23:23
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    @EvanAad ZFC is not complete - remember Godel's incompleteness theorem! Soundness and completeness hold of first-order logic (see Godel's completeness theorem). – Noah Schweber Jun 26 '19 at 23:24
  • @NoahSchweber: Then I don't understand. Henning wrote: "You can take any model of ZFC whatsoever and make it into a model of $\text{ZFC}{\sigma}$ by defining an appropriate interpretation $\sigma$". Assuming ZFC is consistent and has at least one model, this shows that every theorem that can be proved in ZFC can be proved in $\text{ZFC}{\sigma}$. But what about the converse (limited to ZFC forumlas)? – Evan Aad Jun 26 '19 at 23:35
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    @EvanAad No, you've got it backwards. Suppose ZFC$\sigma$ proves $\varphi$. Then if there's a model of ZFC + $\neg\varphi$, that can't be expanded to a model of ZFC$\sigma$. So by contrapositive, if every model of ZFC can be expanded to a model of ZFC$_\sigma$, we get (the strongest kind of) conservativity. – Noah Schweber Jun 26 '19 at 23:37
  • @NoahSchweber: I see. And how do I know that every model for ZFC can be expanded to a model of $\text{ZFC}_{\sigma}$? – Evan Aad Jun 26 '19 at 23:42
  • @EvanAad Just slap on an arbitrary well-ordering for $\sigma$ (here I'm assuming the axiom of choice in the real world, so such a thing exists). Now just check that the remaining axioms hold; this is tedious, but straightforward. The point is that you don't have any axioms which let you get anything from $\sigma$. – Noah Schweber Jun 26 '19 at 23:44
  • @NoahSchweber: OK, but there's the rub: you wrote "I'm assuming the axiom of choice in the real world". Doesn't it take the sting out of the whole thing? I mean, to me it feels very intuitive that every set, or class or any aggregate of things can be chosen from. But then why isn't a global choice axiom a built-in feature of any standard set theory? – Evan Aad Jun 26 '19 at 23:48
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    @EvanAad You seem to be asking two distinct questions in your last comment. RE: "the rub:" ZFC proves that every model of NBG can be expanded to a model of NBG$\sigma$. ZF doesn't. However, since conservativity is an arithmetic principle, by absoluteness we can conclude in ZF alone (indeed in much less) that NBG$\sigma$ is conservative over NBG in the weaker sense. Re: why a selector operator isn't standard: your intuitions aren't necessarily universal, and moreover the study of models where choice fails actually helps with the study of models where choice holds (e.g. in inner model theory). – Noah Schweber Jun 26 '19 at 23:51
  • Regardless, we're now going all over the place. Further questions should be asked as, well, separate questions. – Noah Schweber Jun 26 '19 at 23:53

2 Answers2

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EDIT: There's an issue with defining NBG$_\sigma$: we have to decide whether the original NBG schemes are expanded to apply to formulas involving $\sigma$ too. Otherwise conservativity of NBG$_\sigma$ over NBG is trivial: take a given model of NBG and just "slap on" an arbitrary choice operation. And NBG+ is actually stronger, in terms of $\{\in\}$-theorems, than NBG$_\sigma$.

So - although per the comments below the OP, the weaker version of NBG$_\sigma$ is actually intended - I'm going to say a bit about the stronger version.

In my experience, NBG usually already contains global choice, so NBG=NBG+; but I'll write "NBG" for NBG without global choice below, to match the OP.

Any model $M$ of NBG+ can be turned into a model $\hat{M}$ of NBG$_\sigma$: fix some global choice (class) function $f\in M$ and just name it $\sigma$. Conversely, the reduct of any model of NBG$_\sigma$ to the smaller language of NBG+ is a model of NBG+. So NBG+ and NBG$_\sigma$ prove exactly the same sentences in their common language - that is, the answer to $(1)$ is yes. Note that this implies that $(2)$ and $(3)$ have the same answer.

Let's focus on $(3)$ to ignore the additional symbol. NBG+ already contains a new axiom in the language of NBG, so unless NBG already proves that axiom, the answer will be no.

So we want to find a model of NBG without a global well-ordering. This needs a bit of set theory, but it's doable. If $M$ is a model of ZFC, then we can turn it into a model $N$ of NBG by taking as our classes the definable-with-parameters subsets of $M$. This satisfies global choice iff $M$ has a definable(-with-parameters) well-ordering. And this doesn't necessarily hold. Proving this takes serious work, unfortunately: we show that there is a model of ZFC together with the axiom $$(*)\quad\mbox{"For every $x$, there is some $y$ not in $HOD(x)$."}$$ Here "$HOD(x)$" denotes the class of sets such that they, and each element of their transitive closure, is definable using only ordinals and $x$ as parameters. (HOD stands for "hereditarily ordinal definable," and $HOD(\emptyset)$ is abbreviated "$HOD$.") It's not at all obvious that $(*)$ is consistent with ZFC, or even expressible in ZFC - however, it turns out that both these things hold. Detailed proofs can be found in Kunen's old book.

Noah Schweber
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  • Thanks. You wrote: "In my experience, NBG usually already contains global choice". Could you please point me to some standard reference that corroborates this experience? – Evan Aad Jun 26 '19 at 21:37
  • @EvanAad E.g. nLab. Note that it's not explicitly included there, but it follows from the given axioms (in particular, limitation of size). – Noah Schweber Jun 26 '19 at 21:50
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    It is not clear to me that "the reduct of any model of NBG${}σ$ to the smaller language of NBG+ is a model of NBG+". Note that the new symbol $\sigma$ is defined to be a _function symbol at the first-order-logic level, not a constant that names a class. So a model of NBG${}_\sigma$ might not contain any object that realizes the choice function. – hmakholm left over Monica Jun 26 '19 at 22:56
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    @NoahSchweber: That's what I had a kind of shouting match with the OP about. (For NBG I'm more used to the finite axiomatization presented by Mendelson, where it is clear that one would need an explicit new axiom to allow $\sigma$ in comprehensions -- but that new axiom would essentially be the ordinary axiom of global choice ...). – hmakholm left over Monica Jun 26 '19 at 23:28
  • @HenningMakholm Oh, my apologies! I didn't see that comment thread. Updating my answer now ... – Noah Schweber Jun 26 '19 at 23:28
  • If I understand the first part of your answer correctly, it is possible to infer from it that whenever $\varphi$ is a statement in NBG that is syntactically provable in $\text{NBG}{\sigma}$, it will evaluate to TRUE in every model of NBG. However, is it possible that $\varphi$ is not syntactically provable in NBG? In other words, is it possible for an NBG statement to be syntactically provable in $\text{NBG}{\sigma}$, but not in NBG? – Evan Aad Jun 27 '19 at 06:03
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    @EvanAad No - see Godel's completeness theorem. Being true in every model of $\Gamma$ is the same as being syntactically provable from $\Gamma$. – Noah Schweber Jun 27 '19 at 06:07
  • And this will be the case for any first order theory with a countable number of axioms, of which ZFC, $\text{ZFC}{\sigma}$, NBG, NBG+, and $\text{NBG}{\sigma}$ are all examples, correct? – Evan Aad Jun 27 '19 at 06:09
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    @EvanAad Yes - indeed, any first-order theory with a well-orderable number of axioms (so assuming choice, any theory whatsoever). – Noah Schweber Jun 27 '19 at 06:19
  • Thanks. One more clarification, please. If I understand correctly, the reason why you claimed that the conservativity of $\text{NBG}{\sigma}$ over NBG is trivial, is because, working within ZFC, every standard model of NBG - in the sense that it interprets NBG's '$\in$' as ZFC's $\in$ - can be extended to a model of $\text{NBG}{\sigma}$ by interpreting '$\sigma$' as a choice function (guaranteed by ZFC's axiom of choice) over the set that models the NBG class of sets $V$. However, how do we know that a standard model exists for NBG (or for ZFC, for that matter)? – Evan Aad Jun 27 '19 at 06:39
  • Does every consistent, countable first order theory that is a conservative extension of ZFC admit a standard model? – Evan Aad Jun 27 '19 at 06:39
  • @EvanAad Where did I even mention standard models? The argument works for arbitrary models. (And the answer to your second comment is of course no: consider for example ZFC + "ZFC is inconsistent," which - assuming ZFC itself is consistent, of course - doesn't even have an $\omega$-model.) – Noah Schweber Jun 27 '19 at 06:59
  • Does NBG, at least, admit of a standard model? It is not clear to me how to extend a non-standard model of NBG to $\text{NBG}_{\sigma}$. – Evan Aad Jun 27 '19 at 07:01
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    @Evan: What is "a standard model"? A well-founded one? The existence of well-founded models of set theory has a far greater consistency strength than just "a model". Since NGB and ZFC are equiconsistent, there's no reason to believe that NBG admits a standard model if you meta-theory is just ZFC+Con(ZFC). – Asaf Karagila Jun 27 '19 at 07:18
  • @EvanAad Again, just take any external global selection function and make that your $\sigma$ (which exists, by choice in the ambient universe) - standardness doesn't play a role anywhere. – Noah Schweber Jun 27 '19 at 07:18
  • I think if this is still confusing you you should ask a separate question about it - again, we're getting a bit far afield, and comments aren't really for extended discussion. – Noah Schweber Jun 27 '19 at 07:19
  • @AsafKaragila: By "standard model" I meant a model expressible in ZFC, which interprets NBG's '$\in$' relation as ZFC's $\in$ relation. – Evan Aad Jun 27 '19 at 07:20
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    @Evan: Or equivalently, a well-founded model. (Since those are isomorphic to transitive models with the real $\in$ relation, by Mostowski's collapse lemma.) – Asaf Karagila Jun 27 '19 at 07:21
  • @EvanAad What does "a model expressible in ZFC" even mean? – Noah Schweber Jun 27 '19 at 07:21
  • It means that the foundations in which this model theoretic discussion takes place is ZFC. So we define a set within ZFC theory, which we believe reflects the syntactic NBG theory, and then we define a set within ZFC theory which constitutes a structure, and then we prove by means of ZFC that the structure set is a model of the syntax set. – Evan Aad Jun 27 '19 at 07:29
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    @EvanAad That's not a notion relevant to this discussion - there's no assumption that any of the models here are "definable" in any sense, they are all truly arbitrary - and that's also not at all what the phrase "standard model" means in this context. (And I think this comment thread has hit the law of diminishing returns.) – Noah Schweber Jun 27 '19 at 07:34
  • I may start a new question later today to try to clarify my thoughts and understand this matter thoroughly. – Evan Aad Jun 27 '19 at 07:43
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I will address only question 1. I will take a proof-theoretic approach. I will assume that the axiomatization of the $\text{NBG}$ set theory is the one given here (accessed July 13, 2019), and that the underlying logical framework is first order logic together with the sequent calculus (as described here and here).

Let's consider a fourth theory, which we shall call $\text{NBG}_\tau'$, where $\tau$ is an arbitrary variable of first order logic that does not occur in any axiom of $\text{NBG}$ and that is different than $\sigma$ (note that $\sigma$ is a $1$-place function symbol, whereas $\tau$ is a variable). We define $\text{NBG}_{\tau}'$ by stipulating its set of axioms to be $A \cup \{\varphi\}$, where $A$ is $\text{NBG}$'s set of axioms, and $\varphi$ is the following axiom:

$\tau$ is a function, whose domain is the class of non-empty sets, and such that, for every non-empty set $y$, $\tau(y) \in y$.

Observing that $\text{NBG}_+$'s set of axioms (up to renaming of bound variables) is $A \cup \{\exists\tau(\varphi)\}$, it can be shown on proof-theoretical grounds (see here) that a wff is provable in $\text{NBG}_{\tau}'$ iff it is provable in $\text{NBG}_+$.

Note that $\text{NBG}_{\tau}'$ is similar to $\text{NBG}_{\sigma}$ except that in the former $\tau$ is not a function symbol but rather a set constant assumed to be a function. The question arises as to whether $\text{NBG}_{\sigma}$ and $\text{NBG}_{\tau}'$ are equivalent. We will attempt to answer this question, at least partially. Denote $\text{NBG}$'s set of axioms by $A$, and denote $\text{NBG}_{\sigma}$'s extra axiom by $\xi$.

  1. Every wff that is provable in $\text{NBG}_{\tau}'$ (or, equivalently, in $\text{NBG}_+$) is provable in $\text{NBG}_{\sigma}$. Here is a proof sketch. We wish to show that, for every wff $\psi$ such that $A\cup\{\varphi\}\vdash\psi$ is provable, $A\cup\{\xi\}\vdash\psi$ is provable. By the $(\text{Cut})$ inference rule, it suffices to show that $A\cup\{\xi\}\vdash\varphi$ is provable. Assuming $A\cup\{\xi\}$, define, by the axiom schema for class formation, the relation $\tau := \{(z, \sigma z)\ : z \neq \emptyset\}$, then show that $\varphi$ holds.

  2. We want to show that, for every wff $\psi$ such $A\cup\{\xi\}\vdash \psi$ is provable, there exists some wff $\psi'$ such that the following sequents are provable:

    • $A\cup\{\xi\}\vdash (\psi\iff\psi')$,
    • $A\cup\{\varphi\}\vdash\psi'$.

    I believe this can be shown, but my attempt to do so proved trickier than I expected, so I didn't follow it through.

Evan Aad
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