A question about standard models 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.
 A: 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
A: (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 $y\subseteq v$.
A: 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. 
A: 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.
