Let denote : $\mathcal{L}=\{R\}$ a binary relation, $T_1$ which axiomatizes the class of sets with a number infinite of equivalence classes and each equivalence class contains exactly three elements ($R$ is an equivalence relation). We also have $T_2$ which axiomatises the class of sets with exactly three equivalence classes and each equivalence class contains a number infinite of elements ($R$ is an equivalence relation).

Here are the axioms at the first-order for the equivalence relation :

$\forall x \ R(x,x)$.

$\forall x \forall y \ (R(x,y)\rightarrow R(y,x))$.

$\forall x \forall y \forall z \ ((R(x,y)\wedge R(y,z))\rightarrow R(x,z))$.

For $T_1$:

$\Phi_n : \exists x_1...\exists x_n \ \big(\underset{i\neq j}{\bigwedge \limits_{i=1}^{n}\bigwedge \limits_{j=1}^{n}} \neg R(x_i,x_j) \big)$ for all $n\ge 1$.

$\forall x \exists t \exists u \exists v \ \big(R(x,t)\wedge R(x,u) \wedge R(x,v)\wedge \forall k (R(x,k)\rightarrow(k=t\vee k=u \vee k=v)) \wedge (\neg(t=u)\wedge \neg (t=v) \wedge \neg(u=v))\big)$.

For $T_2$ :

$\exists x \exists y \exists z \ \big(\neg R(x,y) \wedge \neg R(x,z) \wedge \neg R(y,z) \wedge \forall t(R(x,t) \vee R(y,t) \vee R(z,t)) \big)$.

$\Theta_n : \forall x \exists y_1...\exists y_n \ \big(\bigwedge \limits_{i=1}^{n} R(x,y_i) \wedge \underset{i\ne j}{\bigwedge \limits_{i=1}^{n}\bigwedge \limits_{j=1}^{n}} \neg (y_i = y_j)\big)$ for all $n \ge 1$.

Have I write correctly the axioms ?

Now, I think that for $T_2$ a countable model could be $(\mathbb{Z}, \equiv_3)$ but for $T_1$ I don't really see.

Thanks in advance !


1 Answer 1


Your second axiom of $T_1$ means "each equivalence class has at least 3 elements". You must add a condition elements in the equivalence class are all different. Remaining part seems clear.

There is a model of $T_1$: $(\Bbb{Z} \times \{0,1,2\}, \sim)$, where $(a,b)\sim (a',b')$ iff $a=a'$.

  • $\begingroup$ In your example if we take any integer $n$ we have three elements $(n,0), (n,1),(n,2)$ right ? $\endgroup$
    – Maman
    Jan 5, 2017 at 17:27
  • $\begingroup$ @Mamam right, they forms an equivalence class. $\endgroup$
    – Hanul Jeon
    Jan 5, 2017 at 17:31
  • $\begingroup$ Maybe these two theories have $\aleph_0$-categoricity. $\endgroup$
    – Maman
    Jan 5, 2017 at 17:32
  • 1
    $\begingroup$ @Maman They do! $\endgroup$ Jan 5, 2017 at 21:19
  • 1
    $\begingroup$ You could give a back-and-forth argument, but you can also just see it directly... For example, let $M$ and $N$ be countable models of $T_1$. Then $M$ and $N$ each consist of countably many equivalence classes of size $3$. So to give an isomorphism $M\cong N$, you just need to choose a bijection between the classes in $M$ and the classes in $N$, and then choose a bijection between each class in $M$ and its corresponding class in $N$. $\endgroup$ Jan 5, 2017 at 22:37

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