# Can any subset of $\Bbb{N}$ be an equivalence class? [closed]

I am wondering if for any given $$x \in P(\Bbb{N})- \{\emptyset\}$$ we can find an equivalence relation such that it will have an equivalence class equal to $$x$$.

Extend of this question is whether for set $$R$$ of all relations in $$\Bbb{N}$$, the following applies: $$\bigcup_{r \in R} \Bbb{N}/_r = P(\Bbb{N}) - \{\emptyset\}$$

• I don't know what the second part of your question means, but any subset can be an equivalence class. Nov 27, 2022 at 16:02
• @DustanLevenstein Thanks for the answer! Is there perhaps a proof? Secondly, what is unclear in the second part? Nov 27, 2022 at 16:04
• $\{x,\mathbb N - x\}$ Nov 27, 2022 at 16:04
• @LeeMosher fair enough! Nov 27, 2022 at 16:05
• It seems you are thinking that an equivalence relation needs to have a nice description. We see the same for functions because people are used to functions and relations being specified by a description. Lee Mosher's example is a good one partly because it shows no description is necessary. Nov 27, 2022 at 17:35

If you have a subset $$A$$ of $$\mathbb{N}$$, define an equivalence relation $$n \sim m$$ if and only if $$m,n \in A$$ or $$m,n \not \in A$$. It's not hard to show that this is an equivalence relation, and that $$A$$ is an equivalence class of that relation.