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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?

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    $\begingroup$ I assume you mean $\mathcal P(\mathbb N)\setminus\{\emptyset\}$ in the first line? $\endgroup$ – Hagen von Eitzen Nov 8 '14 at 19:00
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Any subset of $\mathbb{N}$ containing $1$ are equivalent. So the equivalence class for $1$ is $$[1] = \{ \ \{ 1 \} \cup A \ \ | \ \ A \in \mathcal P(\{n \in \mathbb{N} \ | \ n > 1 \}) \ \}$$

Hence in general...

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  • $\begingroup$ So there is equivalence class for 2, for 3 and so on? How can I write this down in a correct way? $\endgroup$ – basti12354 Nov 8 '14 at 19:24
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    $\begingroup$ 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 $3, 4, ...$ $\endgroup$ – Simon S Nov 8 '14 at 19:24
  • $\begingroup$ Great explanation thanks. Is there a way to write this down or should I use "..." ;) $\endgroup$ – basti12354 Nov 8 '14 at 19:30
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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.

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