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As I understand it a standard model is a model in which the relation is the $\in$ on the actual set of sets constituting the model.

(i) Hence theories that aren't in the language of set $L_S$ generally won't have a standard model because the binary relation $\in$ can't model binary functions like $+$ in group theory for example. Is this right?

(ii) What do non-standard models of set theory look like? Would someone show me an example? If possible as simple as possible.

Many thanks for your help.

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4 Answers 4

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(1) A standard model of set theory is one in which the membership relation is $\in$; one can have standard models of other theories (e.g., $\Bbb N$ for Peano arithmetic) that don’t say anything about membership.

(2) Let $\varphi(x)$ be $\exists y\big(x=\big\langle y,0\rangle\big)$, and let $\Phi$ be the class of $x$ satisfying $\varphi(x)$. For $x=\big\langle y,0\big\rangle$ and $u=\big\langle v,0\big\rangle$ in $\Phi$ let $x\,E\,u$ iff $y\in v$. More formally, $E$ is the class of $\langle x,u\rangle$ satisfying the formula $\psi(x,u)$ given by

$$\exists y\exists v\left(x=\big\langle y,0\big\rangle\land u=\big\langle v,0\big\rangle\land y\in v\right)\;.$$

Then $\langle\Phi,E\rangle$ is a non-standard class model of set theory that mimics $\langle\mathbf{V},\in\rangle$ in the obvious way. For example, you can check that for $x,u\in\Phi$ as above we have $(x\subseteq u)^\Phi$ iff $y\subseteq v$.

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The book I'm reading defined a standard model as one in which the relation is the actual $\in$. My question is: assuming this definition, can we model other theories? I would think not in most cases. But perhaps in some? I don't think one can model $PA$ using $\in$ but perhaps theories that also only have one binary relation in the signature of their language? –  Matt N. Dec 20 '12 at 9:07
    
I'm aware that it is a stupid question. –  Matt N. Dec 20 '12 at 9:13
    
@Matt: In any model of set theory you can model first-order Peano arithmetic. –  Brian M. Scott Dec 20 '12 at 9:24
    
I'm about to reward two answers. Would you prefer the 50 points of reputation or the accept? –  Matt N. May 5 '13 at 8:26
    
@Matt: It’s your call, but I’m happy to let one of the others have the points. –  Brian M. Scott May 5 '13 at 11:48
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Three quick remarks, rather than a proper answer

1) We certainly talk of standard models of theories other than set theory -- e.g. we talk of standard and non-standard models of first-order Peano arithmetic $PA$. There is an intended interpretation of its language $L_A$, but other ways of interpreting its non-logical vocabulary which still make the axioms of $PA$ come out true.

2) Going right back to Skolem, we know that there must be non-standard countable models of set theory. But there is a limit to how much we can say about them in describing "what they look like". For if I could come up with a nicely detailed story constructing the model and showing that it is a model, I could presumably regiment it in set theory, thereby proving the consistency of ZFC inside ZFC which we know is impossible.

3) Haim Gaifman's paper on the idea of non-standard models is well worth reading: http://www.columbia.edu/~hg17/nonstandard-02-16-04-cls.pdf

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First I should say that I am not a logician so I hope I'm not being totally wrong here. I'm not entirely sure what you mean by the first questions. For the second one, suppose that $M$ is your favourite model for, say, $ZF$. Let's assume that $M$ is standard, so that the elements in $M$ are actual sets and the interpretation of $\in$ is the actual element-hood relation. Let us further assume that $M$ is countable (this is not really needed but you'll see why I add it). Let $f: \mathbb N \to M$ be a bijection.

Define a new model for $ZF$ on $M'=\mathbb N$. Thus, the elements in $M'$ are the 'sets'. Interpret $n\in m$ iff $f(n)\in f(m)$. Clearly, $M'$ is a model of $ZF$ and in it $\in$ is not the actual element-hood for members in $\mathbb N$.

Of course, if $M$ were not countable then you can still do the same trick and re-interpret the model using a bijection between $M$ and any appropriate set of your liking.

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Standard models are models where the real $\in$ is the $\in^M$ of the model. It doesn't have to be that way in general that $\in$ is the interpreted $\in^M$. For example take any model $(M,E)$ and define $M'=M\times\{M\}$ and $(a,M)\mathrel{E'}(b,M)\iff a\mathrel{E} b$. Now we have a model of ZFC whose $\in^{M'}$ is not the real $\in$. If $(M,E)$ was [externally] well-founded then we can collapse it (and we can collapse $(M',E')$ as well) to some $N$ which is a standard model. –  Asaf Karagila Dec 20 '12 at 10:10
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A standard model is often a model which is minimal in a certain sense, like the constructible universe. The interesting question is in what sense it is minimal, because a countable model of ZF would seem to be "more minimal" than the constructible universe. A seemingly total different characterization of standard model from "The seven virtues of simple type theory" just requires that the higher-order axioms (which have been approximated by first order axioms schemes) are true with respect to the standard semantics of higher-order logic. You can imagine this as a condition which maximizes the "dual-space", such that the "primary-space" gets minimized. Or imagine a Galois-connection where "maximizing" one domain will "minimize" the other domain.

Now you may believe that higher-order axioms are devoid of meaning, but this isn't completely true. If you only have a successor function, the first order induction scheme can't "restrict" the model as much as if you also have an addition operator, a multiplication operator, an exponentiation operator, an ... Because higher-order logic allows you to define all these operators and all other "definable" functions and relations, the possible models get "restricted" quite strongly before we even get near to the unclear boundaries of the higher-order axioms.

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Whereas in the context of $\sf PA$ there is a very concrete notion of what is a standard model, and it is indeed the minimal one; in set theory a standard model simply means that its $\in$ relation is the real $\in$ relation (or at least isomorphic to such model). Don't confuse the terminology between two fields of mathematics, even if somehow related. It is true, however, that in a non-standard model there exists non-standard ordinals (i.e. the ordinals of the model are not well-founded) and sometimes its integers are non-standard integers as well. But this is where the similarity ends. –  Asaf Karagila May 12 '13 at 23:14
    
@AsafKaragila Perhaps the answer is not a perfect fit for the question, but it sort of addresses question (i) in my opinion. I intentionally wrote "A standard model is often...", because I also had the impression that the terminology "standard model" is used with slightly different meanings in different fields of mathematics. However, the notion described in my answer has relevance for set theory, both because of Gödel's constructible universe and because of Zermelo's refusal to replace his second order axiom of separation by a first order axiom scheme. –  Thomas Klimpel May 12 '13 at 23:51
    
Do note, however, that there is in fact a minimal standard model (granted these exist to begin with) and it is indeed a model of $V=L$, and it is countable. –  Asaf Karagila May 12 '13 at 23:53
    
@AsafKaragila It's good to know that there exists a countable standard model (granted standard models exists to begin with), and it's even more comforting to know that it can be as nice as the constructible universe (if I understood correctly). I wasn't aware of that. –  Thomas Klimpel May 13 '13 at 0:14
    
Let me clarify my comments. It is consistent that there are no models, and it is consistent that there are models, but no standard models. But if there are standard models, then there are countable standard models, in which case there is a minimal standard model, and that minimal model satisfies $V=L$. –  Asaf Karagila May 13 '13 at 0:19
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