I got the set:

$$ M=\{1,2,3,4\}. $$

I could split the power set of M into the following subsets:

$$ P_{0}=\{\emptyset\} \\ P_{1}=\{\{1\},\{2\},\{3\},\{4\}\} \\ P_{2}=\{\{1,2\},\{1,3\},\{1,4\},\{2,3\},\{2,4\},\{3,4\}\} \\ P_{3}=\{\{1,2,3\},\{1,2,4\},\{1,3,4\},\{2,3,4\}\} \\ P_{4}=\{\{1,2,3,4\}\} $$

The set of those subsets would give me a partition of the power set:

$$ P=\{P_{0},P_{1},P_{2},P_{3},P_{4}\} \\ $$

We can interpret the partition as a set of equivalence classes. The equvalence relation would be then defined as:

$$ aRb :\Leftrightarrow |a|=|b| $$

Is this correct?

  • 1
    $\begingroup$ Yes. You can test for yourself that this is an equivalence relation by finding if it is reflexive, symmetric, and transitive. $\endgroup$
    – anak
    Jul 24, 2015 at 15:33

1 Answer 1


Yes, that's correct. By your definition two elements a and b (elements i.e. subsets) are equivalent if their cardinality is the same.

  • $\begingroup$ Exactly. Would there be another way to define the relation, so I would get the desired classes? $\endgroup$
    – craaaft
    Jul 24, 2015 at 15:36
  • $\begingroup$ Well, anything which you can express formally and which means (logically) the same thing is OK. I cannot think of another way right now but there may be another way. $\endgroup$ Jul 24, 2015 at 15:42
  • $\begingroup$ OK. Thank you very much! $\endgroup$
    – craaaft
    Jul 24, 2015 at 15:51

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