# Equivalence class question with the following

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?

• Please don’t post pictures, and definitely don’t ask people to leave the site to look at a scan. – Arturo Magidin May 27 at 23:40
• 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... – JMoravitz May 27 at 23:49

## 1 Answer

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 $$n\in \mathbb{Z}.$$