So I want to practice with understanding the definition of Spectra. Basically, I understand it is the set consisting of all natural numbers n such that there is a model of phi with exactly n elements.
1) If we let L = {E} and let phi be a sentence which says that E is an equivalence relation in which every class has precisely three elements, how can we show that the spectrum of phi is the set of naturals which are divisible by 3?
Writing it out, I'm thinking to let E = {E| a,b,c are Naturals} and I dont see how the divisible by 3 fits in
I’ll do the first one completely to put you on the right track and let you think a bit more about the second.
Let $\varphi$ be the following sentence:
$$\begin{align*} \forall x\,E(x,x)&\land\forall x,y\big(E(x,y)\to E(y,x)\big)\\ &\land\forall x,y,z\big(E(x,y)\land E(y,z)\to E(x,z)\big)\;; \end{align*}$$
it says that $E$ is an equivalence relation. The first conjunct says that $E$ is reflexive; the second, that $E$ is symmetric; and the third, that $E$ is transitive. Now let $\psi$ be the following sentence:
$$\forall x\exists y,z\Big(E(x,y)\land E(x,z)\land x\ne y\land x\ne z\land y\ne z\land\forall w\big(E(x,w)\to(w=x\lor w=y\lor w=z\big)\Big)\;,$$
it says that each element is related by $E$ to exactly two other elements. (Make sure that you see why this is so.) Thus, $\varphi\land\psi$ is a sentence saying that $E$ is an equivalence relation such every equivalence class of $E$ has exactly three elements. If $\mathfrak A$ is a finite model of $\varphi\land\psi$ with $k$ equivalence classes, then $|\mathfrak A|=3k$. Clearly we can define such a model for each $k\in\Bbb N$, so the spectrum of $\varphi\land\psi$ is $\{3k:k\in\Bbb N\}$, the set of non-negative multiples of $3$.