# Finding the equivalence class of this relation

I am having this relation:

$$A=\mathcal P(\mathbb {N} \diagdown 0) ,$$

                             A~B :<=>  min A = min B


I haved already proved, that it is a equivalence relation. Now I have to find an equivalence class for this relation. I am knowing the definition of an equivalence class $$[x] := y \in A|\ xRy$$

What is the amount of the equivalence classes of this relation?

• I assume you mean $\mathcal P(\mathbb N)\setminus\{\emptyset\}$ in the first line? – Hagen von Eitzen Nov 8 '14 at 19:00

Any subset of $\mathbb{N}$ containing $1$ are equivalent. So the equivalence class for $1$ is $$[1] = \{ \ \{ 1 \} \cup A \ \ | \ \ A \in \mathcal P(\{n \in \mathbb{N} \ | \ n > 1 \}) \ \}$$
• Yes. Any subset of $\mathbb{N}$ containing $2$ but not $1$. You can model an explicit expression for $[2]$ on the one above. Similarly for $3, 4, ...$ – Simon S Nov 8 '14 at 19:24
Note that \begin{align}f\colon \mathcal P(\mathbb N)\setminus\{\emptyset\}&\to \mathbb N\\S&\mapsto \min S\end{align} is onto and for each $n\in\mathbb N$, the preimage $f^{-1}(n)$ is an equivalence class.