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For $A = \{a, b\}$ and $B = A^* × A^*$ ($A^*$ is the set of all words (of all lengths))

In my opinion, because $B$ is reflexive, symmetric and transitive $(aa, bb)$. It is an equivalence relation, the equivalence classes should be one, because all elements are related to each other?

Am I correct?

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  • $\begingroup$ What is $B$ a relation on? Seemingly, it's on the set of all words with letters in $A$. $\endgroup$ – астон вілла олоф мэллбэрг Aug 18 '17 at 5:55
  • $\begingroup$ That's right! A* could be any length. $\endgroup$ – user3741679 Aug 18 '17 at 6:12
  • $\begingroup$ Okay, so $B$ is a relation on the set $A^*$, given by $A^* \times A^*$ itself. Of course, this means any two elements are related, so that the number of equivalence classes is indeed only one. $\endgroup$ – астон вілла олоф мэллбэрг Aug 18 '17 at 6:16
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Yes you're right. Two elements belong to the same equivalence class if and only if they are equivalent. If you take any two elements $w_1,w_2\in A^*$, we have $(w_1,w_2)\in A^*\times A^*=B$, which shows that they are both equivalent under the equivalence relation $B$.

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