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  1. Let $A=P(\Bbb Z)$

    • Prove that $R=\{(S,T)\in A{\times}A: \exists n\in \Bbb Z~\forall x\in \Bbb Z~(x\in S\leftrightarrow x+n\in T)\}$ is an equivalence relation on $A$ (15 points).

    • Write down all elements in $A/R$ that consist of a single element (7 point)

I already proved that $R$ is an equivalence relation, but I am unsure about the second question regarding the equivalence class. If $S$ and $T$ were both empty sets, would that be the equivalence class that contained only one element?

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    $\begingroup$ Please don’t post pictures, and definitely don’t ask people to leave the site to look at a scan. $\endgroup$ – Arturo Magidin May 27 at 23:40
  • $\begingroup$ Rephrased in a bit more plain of English, your equivalence relation $R$ can be phrased as "$S$ is related to $T$ iff $T$ is just $S$ where each element has been 'shifted' by some number." For example $\{2,3\}$ is related to $\{7,8\}$ since the second is just the first but with each number shifted up by $5$. On the other hand $\{2,3\}$ is not related to $\{10,20\}$. You should be able to convince yourself that any set with at least one element is related to infinitely many other sets. Eg, In the case of $\{2,3\}$ it is related to each of $\{3,4\},\{4,5\},\{5,6\},\dots,\{1000,1001\}$ etc... $\endgroup$ – JMoravitz May 27 at 23:49
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Well, $\emptyset\, \mathcal{R}\, \emptyset$ and $[\emptyset]_{\mathcal{R}}=\{\emptyset\}$ hence the equivalence class of the empty set indeed contains only one element (in fact, it's the only one). Also, notice that if $S\,\mathcal{R}\,T,$ then $\#(S)=\#(T)$ - in fact, there is a bijection between the elements of $S$ and the elements of $T$ (and this is why the equivalence class of the empty set contains only the empty set itself). How can you use this fact to conclude?

Hint: for any non-empty set $S,$ consider the sets $T_n=\{s+n: s\in S\}$ and prove that $S\,\mathcal{R}\,T_n$ for any $n\in \mathbb{Z}.$

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