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I'm new to model theory/set theory. I'm confused by the following definition from the book "Fundamentals of Model Theory" by W. Weiss & C.Mello:

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If I understand well, the consequence of this definition is that to prove a theorem of a theory (for example ZFC), we have to prove its exactitude for EVERY model of ZFC. However, it seems to me that we don't do it at all, we prove any theorem by using the framework of 1 model : (class of all sets, $\in$ relation) and we claim that it is a theorem of the theory. This doesn't satisfy the above definition.

Could you please tell me if my reasoning is correct or not ? Are what we usually claim "theorems" in ZFC really theorems when we haven't check it for all models of ZFC (which seems to be impossible) ? Or maybe there is a flawness in the definition ? Thanks a lot.

VDT-QHH
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  • Usually, we prove a theorem deriving it by rules of logic from axioms of the theory. This means that, if $T$ is the collection of axioms and $\sigma$ is the theorem, we have $T \vdash \sigma$. By completeness of the underlying logic, we have $T \vDash \sigma$ that means that sentence $\sigma$ (the theorem) is true in every model of the axioms $T$ (the theory). – Mauro ALLEGRANZA Apr 24 '24 at 09:01
  • @MauroALLEGRANZA: Hello, actually it seems to me that we prove a theorem deriving it by rules of logic using the "interpretations" of the axioms of the theory within a model. For example I see that in set theory, I see that we reason directly by using the notion of "sets" and $\in$ relation (with precise meaning, not just a relation symbol of the language). In fact, I can't imagine how could we prove things by just using the language of the theory without applying it to the model. But if it is the case, then I think that all we can say is that "there is 1 model satisfies the theorem"... – VDT-QHH Apr 24 '24 at 09:07
  • You have to "complete" the Definition 12 (page 12) of theory and consequence and the Proposition of page 14: The Completeness Theorem (Godel, Malcev): "A set of sentences is consistent if and only if it is satisable" with the general form of the Soundness and Completeness Theorem: $T \vDash s$ iff $T \vdash s$. – Mauro ALLEGRANZA Apr 24 '24 at 09:10
  • You can see here some examples of theorems of set theory. – Mauro ALLEGRANZA Apr 24 '24 at 09:12
  • In general, see e.g. Goldrei, Set Theory. – Mauro ALLEGRANZA Apr 24 '24 at 09:14
  • @MauroALLEGRANZA: Hello, thanks a lot. I have seen your answer of the proof of theorem using formal language. It is impressive yet tired. I think that isn't usually the case where we conduct a proof. I insist on the fact that we usually prove things inside a model. Do you agree with me on this ? From the general form of Soundness and Completeness Theorem in Wiki, I see that we still have to prove things (to have $T⊢s$) using directly formal language of a theory. Do you know if there exists anything "If a theorem is satisfied for a model, then it is satisfied for all models of the theory?" – VDT-QHH Apr 24 '24 at 09:37

3 Answers3

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If you want to directly check the definition, then yes, this is what you have to do.

In practice, we sometimes do just that, but typically, to prove that a given sentence $\sigma$ is a consequence of $T$, we can also do one of the following:

  • We show that $T\vdash \sigma$, i.e. $\sigma$ is a theorem of $T$; to do this, we only need to find a proof, which is a purely syntactic thing and does not say anything about any models; this is enough, because by the soundess theorem, $\vdash$ implies $\models$ (the converse is also true by the completeness theorem). For example, to show (in ZFC) that given sets $a,b,c$ there is a set whose elements are precisely $a,b$, and $c$, you can simply apply the axiom of pairing three times and then apply the axiom of union, arguing that $\{a,b,c\}=\bigcup\{\{a,b\},\{c,c\}\}$.
  • If $T$ is complete (for instance, it is a theory of a model), then $T\models \sigma$ if and only if $M\models \sigma$ for some model $M$ of $T$. Then we can simply pick a model that is convenient to work with. For instance, this is a typical way of showing theorems in real closed fields: we show that they hold in real numbers, where we can use the full power of real and complex analysis, and it follows that they hold in an arbitrary RCF.

In your example of ZFC, the theory is not complete, and models are not really accessible, so basically, we follow the first approach.

Mind, the actual proofs of theorems of ZFC or other set theories can often be quite a bit more complicated, especially when you use (an extension of) ZFC as the metatheory (which you often do) and start working with forcing extensions and whatnot. However, the formal intricacies of proofs in axiomatic set theory are beyond the scope of this question.

tomasz
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  • Hello, thanks!! For this part of the first bullet point: "then $T⊨σ$ if and only if for some model of $T$", it seems to me that the statement is not complete... In fact, I understand your first bullet point but I think that from the moment we talk about "set", $\in$, I think we actually choose 1 model of ZFC as the framework to conduct the proof. MauroALLEGRANZA has stated a link to his proof of 1 theorem of ZFC but I think that it is pratically impossible to reproduce that proof for a more complex theorem. So i think even in ZFC, we are using the second approach.... – VDT-QHH Apr 24 '24 at 09:47
  • ... But then in this case, we can't be sure that the theorem is correct for other model of ZFC. So basically, the theory and what I see in practice confuse me. Ref to the proof of MauroALLEGRANZA : https://math.stackexchange.com/questions/2815079/how-to-prove-statements-in-formal-set-theory-substitution-the-empty-set-and-an – VDT-QHH Apr 24 '24 at 09:48
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    Yes, I missed a piece, sorry, it's fixed now. No, we are not using the second approach, because ZFC is not complete. We do not need to practically reproduce a formal proof (although to my understanding, this is done to some degree for the purpose of automatic theorem proving) verbatim in the language of set theory. Instead, we write a proof in an informal language in a way that can in principle be easily (if tediously) converted to a formal proof, much like I did in the first bullet. – tomasz Apr 24 '24 at 09:59
  • Hey, thanks! it is clearer for me now. Concerning what you said about the real closed field, I'm a little bit confused because as I understand, ZFC lays the foundation for others mathematical objects: we defined natural numbers, real numbers, etc on the basis of ZFC. So ultimately, I think the theorem about real numbers/real closed field is also a theorem of ZFC (which isn't a complete theory). Therefore, a theorem is true in a model isn't necessarily true in another model... Maybe what you mean is that the theory of Real closed field has nothing to do with ZFC (I doubt this) ? – VDT-QHH Apr 24 '24 at 10:10
  • If you consider ZFC as the foundational theory of mathematics, then all mathematical theorems are theorems of ZFC, so what you say is sort of vacuously true in that way. However, most mathematical ideas are not actually based on ZFC, it is merely used as a convenient way to lay a formal foundation where necessary. In particular, "natural" properties of the field of real numbers (and most would agree that the first order properties are natural) should not depend on the foundations. – tomasz Apr 24 '24 at 10:19
  • In particular, formal proofs of such theorems should be reproducible in any reasonable foundational metatheory (otherwise, it wouldn't be a good metatheory). The details and various sides of all this lie within the realm of foundations of mathematics, not model theory, and are frankly more philosophical than mathematical (and beyond the scope of this question). – tomasz Apr 24 '24 at 10:21
  • So basically yes, I am saying that the theory of real closed fields (and most "natural" theories) has (little to) nothing to do with ZFC. The same cannot be said about many of their models, which are often built using set-theoretic tools (but again, arguably, "meaningful" things should arguably not depend on particular foundations). – tomasz Apr 24 '24 at 10:29
  • Do you think it is reasonable for me to think of the theory of RCF like this: this theory has its own language (which is different from the language of ZFC), and given a model of this theory, there is the notion of "real numbers", in this case, "real numbers" should be understood as a primitive object rather than questioning "what is a real number ? how can we construct it?". So basically, what i mean is that is it better not to say that real numbers in RCF theory are "indeed" the real numbers we construct within the standard model of ZFC ? (i.e. we study RCF on its own right) – VDT-QHH Apr 24 '24 at 12:16
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    @VDT-QHH: Yes, that seems sensible to me, as long as you keep the quotes around "real numbers". :-) – tomasz Apr 24 '24 at 20:36
  • Hello, I have a small question please. Do you have the name of the theorem that claims "If $T$ is complete (for instance, it is a theory of a model), then $T⊨σ$ if and only if $M⊨σ$ for some model $M$ of $T$" as you said in your answer? I thought it is completeness theorem but in fact, when reading the theorem again, i realize that it doesn't say anything about "if the theorem is true for 1 model, then it is true for all models of the theory". Thanks for your help! – VDT-QHH Apr 26 '24 at 07:28
  • Moreover, in your comment above, you said that "The same cannot be said about many of their models, which are often built using set-theoretic tools", could you please provide me some examples (names) of these model ? I am thinking of measure theory which is built based on set theory. – VDT-QHH Apr 26 '24 at 09:51
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    I don't think it has a name, maybe except for "law of excluded middle" (plus soundness). If $M\models T$ and $M\models \sigma$, then $T\not\models\neg\sigma$, so $T\not\vdash \neg\sigma$, so by definition, if $T$ is complete, then $T\vdash\sigma$, whence $T\models\sigma$. – tomasz Apr 26 '24 at 16:24
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    Regarding your other comment, I don't think it makes sense to talk about names. The point is that models are sets, so we often use set-theoretic concepts to build them. This is (necessarily, I think) quite vague as to what kind of model is considered natural, and which is an artifact of set theory. – tomasz Apr 26 '24 at 16:27
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If we have a statement $\sigma$ that we wish to prove in some theory, then usually, the way that we do that is by using the axioms of that theory and the inference rules of our logic system. Once we have done that, we then know that $\sigma$ is true in all models of the theory. The fact that it's true in all models is a consequence of the proof, not a prerequisite for the proof.

You ask: isn't it true that we usually prove things inside a model? The answer is no, absolutely not! In particular, when we prove theorems of ZFC, we almost never have a particular model of ZFC in mind. Theoretically, we are merely applying axioms and inference rules, and there are no models anywhere to be seen.

It is true that when we work in ZFC, we typically imagine that there is some particular collection of objects, $V$, that the axioms are "talking about." However, this $V$ that we have in mind is usually not actually a model of ZFC. It's just an abstract, fictional idea that we use as a mental aid.

In particular, "(class of all sets, $\in$ relation)" is not a model of ZFC. If we already have some model $\mathcal{M}$, then we may use the phrase "the class of all sets of $\mathcal{M}$" to refer to one of the components of $\mathcal{M}$, but the phrase "the class of all sets" all by itself doesn't give us a model.

Admittedly, some mathematicians believe that there actually is one particular model of ZFC that is the "correct" or "intended" model. However, even if we really do have a particular model of ZFC in mind, we nevertheless have to obey a restriction: when we prove things, we're not allowed to use arbitrary properties of the model in our proofs. Instead, we must only use those properties which are axioms of ZFC. By doing this, we guarantee that our proofs hold true in all models of ZFC, not only the particular model we had in mind.

Sophie Swett
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  • Hello, thanks for your answer! I'm a little bit confused because what you said in the 1st and 2nd paragraph is in some sense not the same thing as the second bullet point of the answer of tomasz, which I am more comfortable. Could you please share with me your thoughts on this ? – VDT-QHH Apr 24 '24 at 13:32
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    @VDT-QHH Let me start by paraphrasing the bullet point you're referring to. If I understand correctly, tomasz is saying that if a theory is "complete," then we actually can just pick one model of that theory and prove things about that model, and the results will be true for all models of that theory. Importantly, however, ZFC is not a "complete" theory, so this approach can't be used for ZFC. – Sophie Swett Apr 24 '24 at 17:59
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    @VDT-QHH It would probably be valuable to try to explain exactly why you think that "we usually prove things inside a model." Maybe your idea of what the word "model" means is different from mine. In particular, I don't know whether or not there are any models of ZFC; it seems like it would be difficult for me to pick a model of ZFC when I don't know whether such things exist or not. – Sophie Swett Apr 24 '24 at 18:04
  • Hello, what i mean is that I usually see we prove theorems by using the interpretation of the language in a given model (for example in ZFC, we reason by using the notion of "set" which is not a precise notion in the language of ZFC), or as tomasz said, we prove the theorems of real closed field using the model without passing through the language of that model. – VDT-QHH Apr 24 '24 at 20:33
  • @VDT-QHH May I ask exactly what you mean when you say "model"? It sounds like you're thinking of the word "model" as meaning something along the lines of "collection of concepts that the symbols in the theory are considered to represent." If that were what the word "model" means, then I think everything you've said would make perfect sense. However, the word "model" actually means something very, very different from that, and that may be the source of your confusion. – Sophie Swett Apr 25 '24 at 03:13
  • For me, a model is a structure (domain of discourse, set of interpretation functions, relations, constants of the symbol of the language), so basically what you meant I think. But if what i said makes sense, then following your answer, to prove a theorem of a theory, we will always use the language of the theory for the proof. But as tomasz has stated, we can also prove that the theorem is true in a specified model, then if the theory is complete, then the theorem is also true in another model of the theory. So i'm confused between your answer and what you said in the comment above. – VDT-QHH Apr 25 '24 at 06:17
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    @VDT-QHH Well, what I wrote in my answer is that if we want to prove a theorem, we will usually use the axioms of the theory for the proof. It's true that for complete theories, you can use that "alternative method" of proving the theorem for a particular model. However, it seems like you're mainly interested in incomplete theories (such as ZFC), and this "alternative method" doesn't work for those. That's why I didn't mention the method in my answer. – Sophie Swett Apr 25 '24 at 13:29
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I think the existing answers, while correct, overcomplicate things.

Think of the following analogous question:

To prove a theorem [about natural numbers], we have to prove its exactitude for EVERY [natural number]. However, it seems to me that we don't do it at all, we prove any theorem by using the framework of 1 [natural number $x$], and we claim that it is a theorem of the theory.

Of course the response to this question is clear: there is no "specific object $x$." Instead, $x$ is a placeholder for an "arbitrary natural number" (which is admittedly a bit of a subtle notion) and we're being careful to only use facts true about every natural number when reasoning about $x$.

Exactly the same thing is going on here. When we prove "$\varphi$ is a theorem of $\mathsf{ZFC}$," what we do is:

  • argue that $\varphi$ is true in $V$,

  • where the only things we use about $V$ are that it satisfies the $\mathsf{ZFC}$ axioms.

Depending on the reader, there is here the additional mental baggage that we really do have a specific model in mind, namely the "true universe." But the point is that we are careful to only use the $\mathsf{ZFC}$-satisfying-ness of the model we reason about.

Noah Schweber
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  • Hi Noah, thanks for your answer! However, I think what you meant is equivalent to the fact that we prove a theorem of $ZFC$ using directly the formal language/axioms of $ZFC$ except for we have a "imaginary" model in mind to help with the proof. – VDT-QHH May 01 '24 at 08:40