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I have the following equivalence relation problem.

$Let \ R\subseteq 2^S*2^S = \{(A,B):|A\cap T|=|B \cap T|\}\ where\ S=\{0,1,2\} \ and \ T=\{1,2\} \ Show \ that \ R \ is \ a \ equivalence \ relation \ to \ 2^S \ and \ describe \ the \ equivalence \ classes\ $

I understand that in order to show that $R$ is an equivalence relation I need to show that is $R$ is symmetric, reflexive and transitive.

Though I have two questions.

  1. What the result of $2^S$ will be?
  2. What steps do I need to follow to find the equivalence classes of an equivalence relation?
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  • $\begingroup$ I think you meant $\;2^s= P(S)=$ the set of all subsets of $\;S\;$, and then $\;R\subset 2^S\times 2^S\;$ . What you mean by (1) is beyond my understanding, and about (2): there are only $\;8\;$ elements in $\;2^S\;$ so it shouldn't be that hard to do the actual partition of $\;2^S\;$ by equivalence classes. For example, $\;[\emptyset]=\{\emptyset\,,\,\{2\}\}\;$ $\endgroup$ – Timbuc Feb 8 '15 at 13:34
  • $\begingroup$ I thought 2^S to be power. I guess I missed the symbolism of the power Set. 1 is now understood. About 2 I still have a hard time understanding how you found [0]={0,{2}} $\endgroup$ – Dimitri C Feb 8 '15 at 13:38
  • $\begingroup$ Nice, now try to separate all the elements in $\;2^S=P(S)\;$ in different equivalence classes. $\endgroup$ – Timbuc Feb 8 '15 at 13:40
  • $\begingroup$ How can I do that? $\endgroup$ – Dimitri C Feb 8 '15 at 13:46
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I already gave you an example in the comment. For another one:

$$|\{1\}\cap T|=1\implies X\in\left[\,\{1\}\,\right]\iff |X\cap T|=1$$

so for example, we have in this case

$$\{1\}\,,\,\{0\}\,,\,\{0,2\}\,,\,\{1,2\}\in\left[\,\{1\}\,\right]$$

Now you try other cases.

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  • $\begingroup$ I think I got it. Will the answer(to my problem) be [0]={{e},{0}} [1]={{1},{2},{0,1},{0,2}} [2]={{1,2},{0,1,2}} ? $\endgroup$ – Dimitri C Feb 8 '15 at 13:55
  • $\begingroup$ Observe that $\;[1]=[2]\;$ so your last one cannot be right, and in fact observe that $$|\{0,1,2\}\cap T|=2\neq 1=|\{1,2\}\cap T|$$ $\endgroup$ – Timbuc Feb 8 '15 at 14:58
  • $\begingroup$ You are right. T is actually {1,2} and not {0,1} . $\endgroup$ – Dimitri C Feb 8 '15 at 15:02

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