-2
$\begingroup$

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: $$\bigcup_{r \in R} \Bbb{N}/_r = P(\Bbb{N}) - \{\emptyset\}$$

$\endgroup$
5
  • 1
    $\begingroup$ I don't know what the second part of your question means, but any subset can be an equivalence class. $\endgroup$ Nov 27, 2022 at 16:02
  • $\begingroup$ @DustanLevenstein Thanks for the answer! Is there perhaps a proof? Secondly, what is unclear in the second part? $\endgroup$
    – Stanley
    Nov 27, 2022 at 16:04
  • 4
    $\begingroup$ $\{x,\mathbb N - x\}$ $\endgroup$
    – Lee Mosher
    Nov 27, 2022 at 16:04
  • $\begingroup$ @LeeMosher fair enough! $\endgroup$
    – Stanley
    Nov 27, 2022 at 16:05
  • $\begingroup$ 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. $\endgroup$ Nov 27, 2022 at 17:35

1 Answer 1

3
$\begingroup$

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.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged .