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I need to find the elements of set $A$, and the relation $R$ given its quotient set $A/R = \{\{c\}, \{a,e\}, \{b,f,g\}\} $

I think I can deduce the elements of $A$ as being $A=\{a,b,c,e,f,g\}$.

However, how can I reconstruct the initial relation? Taking a look at the quotient set, I could deduce that we have three equivalence classes. If we take the first one, say $ \{c\}$:

By definition, for an equivalence class, we have that $$[a]_R = \{b | b \in A, a \sim_Rb\}$$

This would mean that one element from set $A$ is only in relation with $c$. My main question is, how can I know which exact element that is?

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  • $\begingroup$ If $\{c\}$ is an equivalence class then it means that $c$ is equivalent only to itself. By the way, you missed the element $e$ at the beginning. $\endgroup$
    – Mark
    May 31 at 16:07
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    $\begingroup$ To be clear... the set of equivalence classes itself is enough information to define the relation. $x$ is related to $y$ iff $x$ and $y$ are both members of the same equivalence class. That said, there doesn't need to be a convenient way to describe the relation without referring to the equivalence classes. $\endgroup$
    – JMoravitz
    May 31 at 16:12
  • $\begingroup$ Sure, we could come up with some contrived example like "two elements are related if they are both vowels or if they are both c, or are both consonants which are not c" but that doesn't exactly roll off the tongue here and doesn't make things easier to read. $\endgroup$
    – JMoravitz
    May 31 at 16:13
  • $\begingroup$ @JMoravitz how? Suppose I have A={a,b,c,d}, R={(a,b), (a,c), (b,c), (b,d)}. The equivalence class of [b] = {c, d}, but we don't have a pair (c,d) in the relation R? $\endgroup$
    – THE_CRANIUM
    May 31 at 16:19
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    $\begingroup$ @THE_CRANIUM that choice of $R$ in your comment is not an equivalence relation and as such does not have equivalence classes. $\endgroup$
    – JMoravitz
    May 31 at 16:37

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