Let's say the model $\alpha$ consists of $(A, a(R_1), a(R_2)...a(R_n))$, where A is a finite set, and all $R_i$ are relation symbols, $a$ denotes their interpretation in the model. I need to prove that a sentence $\phi_a$ exists, which satisfies the following for any model $\beta$:

$$\beta \models \phi_a \iff \beta \cong \alpha $$

Also, let's denote the elements of A as $a_o, a_1... a_n$

The guide says to start by constructing a formula $\psi$, telling which relation each string of elements belongs to (using free variables $v_o... v_n$ and quantifiers). But I am not exactly sure how to do this and how it relates to the question.


Write down a sentence that says: there are $n$ objects and only $n$ objects, and those objects fulfill the relation $R_1$ in this particular way, and $R_2$ in that particular way, and so on.

One can write a sentence like that without too much effort. It will start like this $$\exists x_1\ \exists x_2\ \cdots\ \exists x_n [\ \ \dots\dots\dots\dots \ \ ],$$ where inside the brackets we say, first that every $z$ is equal to one of the $x_i$'s, and all the $x_i$'s are different, and then we also say things like $R_1(x_2,x_5)$ and $R_1(x_3,x_2)$ and so on, for all the instances that hold in the original model, and then we say things like $\neg R_1(x_4,x_7)$ and so on for the instances that don't hold. Do you see how it goes?

The thing to notice is that any model of this sentence will look just like your given model (except that I used $n$ instead of $n+1$), and they will be isomorphic. The way that the model $\beta$ realizes the existential quantifiers will provide the isomorphism.

  • $\begingroup$ Thank you, is it normal to assume that the relation symbol takes two arguments? That would simplify writing down the formula quite a bit for sure... $\endgroup$ – Tony Feb 4 '17 at 3:34
  • $\begingroup$ I just meant to give an example. If you have a trinary relation, you would use $R_1(x_2,x_1,x_5)$ or whatever. It is the same idea. The point is that in principle, one could write down a complete description of the model this way. (You don't have to actually write it down, but just prove that one could do it, in principle.) $\endgroup$ – JDH Feb 4 '17 at 3:40
  • $\begingroup$ I am still having some trouble seeing why the model has to be isomorphic if the formula is true in model $\beta$. Primarily, it seems as though I can have more than n elements in the model $\beta$ and thus it's possible to not have bijection between the models. How does the formula prevent such cases? $\endgroup$ – Tony Feb 4 '17 at 11:20
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    $\begingroup$ @Tony I had mentioned that the formula says, "there are $n$ and only $n$ objects." Inside the brackets, I said it says, "every $z$ is equal to one of the $x_i$ and the $x_i$ are all different." So every model of the sentence must have exactly $n$ objects. $\endgroup$ – JDH Feb 4 '17 at 12:22

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