# Sentence equivalent to $\bigwedge_{i=1}^ \infty \sigma_i$ without using infinite conjunctions

Let $$\mathcal{L}$$ be a language (with equality) containing a binary relation symbol $$R$$.

Assume $$R$$ is interpreted as an equivalence relation (so it is reflexive, symmetric, and transitive)

I see that the sentence $$\sigma_1$$: $$\exists! x \forall y(xRy \to x=y)$$ says that there is a unique equivalence class of size $$1$$. Similarly, we can define $$\sigma_2$$ by $$\exists^{=2}x \exists!y(x \neq y \wedge \forall z(zRx \to z=x \vee z=y)$$, which says there is a unique equivalence class of size $$2$$. (Here, the notation $$\exists^{=2}$$ is an abbreviation for "there exist exactly two" in the obvious sense.)

Similarly, I see that, in principle, one could define $$\sigma_k$$ be the sentence saying that there is a unique equivalence class of size $$k$$.

Q: How can we express that there is a unique equivalence class of size $$k$$ for each $$k$$?

If infinite conjunctions were allowed, I would just write $$\bigwedge_{i=1}^ \infty \sigma_i$$ and call it a day. However, they are not.

I was thinking perhaps of adding a sentence $$\sigma$$ which says "for each equivalence class, there exists a unique equivalence class of size $$1$$ larger." Then I would write down $$\sigma_1 \wedge \sigma$$ and call it a day. The only problem is that I don't know how to express $$\sigma$$ formally. (Is it even possible to express it formally?)

Please let me know if there are any typos in the above.

Let $$T=\{\sigma_k:k\in\omega\}$$.
Suppose there is a formula $$\sigma$$ equivalent to $$T$$. In partucular $$T\vdash\sigma$$. By compactness a finite subset $$T_0\subseteq T$$ implies $$\sigma$$, a contradiction.