This question arose from the question: Intersection of equivalence classes of two equivalence relations

This is the initial task, same as on the link above:

Let $A$ be a nonempty set and $\sim$ and $\thickapprox$ two equivalence relations on the set $A$.

Relation $\triangle$ is defined like this: $x,y\in A,\;x\;\triangle\;y\;\iff x\;\sim\;y\;\wedge\;x\;\thickapprox\;y.$

Prove these statements:

$1)$ $\triangle\; $ is an equivalence relation on the set A.

$2)$$P\in A_{/\triangle}\iff \exists \;Q\in A_{/\sim}\;\wedge\;R\in A_{/\thickapprox}\;\;P=Q\cap R$

After examining comments and suggestions, I tried this:

If a relation of equivalence generates the quotient set (please correct me if it is false) and classes of equivalence are its elements, can we find two different partitions and their possible intersection that would also be an equivalence class to prove the statement no. $2$?

  • $\begingroup$ The expression a "relation of equivalence " means nothing. $\endgroup$ – William Elliot Nov 12 '19 at 21:38
  • $\begingroup$ @WilliamElliot, when I first saw your comment, I checked if I' d made some kind of semantic mistake in translation because in my mandatory literature it is always in the genitive case, but when I searched for both of the expressions, there were many results reffering to each of them. This expression might have been used by people who aren' t native English speakers. Even in the previous exams it always says, for example: 'Check if ρ is a relation of equivalence'. What you're saying is a good critic because I wasn't consistent in using one term. $\endgroup$ – Praskovya2.718281828 Nov 12 '19 at 22:11
  • $\begingroup$ @WilliamElliot Anyway, thank you for commenting. It is also a part of learning for me as a 'foreign' student $\endgroup$ – Praskovya2.718281828 Nov 12 '19 at 22:14
  • $\begingroup$ I disagree with William Elliot that "relation of equivalence" is not meaningful in English. However, it does sound archaic - a construction that just isn't used in modern English. On the other hand, "Classes of equivalence" is wrong even by archaic English grammar (at least to my knowledge). The difference is that "classes" have members, and "of" indicates membership, so "classes of equivalence" would indicate the members of the class are equivalences themselves, instead of objects that are equivalent to each other. $\endgroup$ – Paul Sinclair Nov 13 '19 at 3:26
  • 1
    $\begingroup$ Excellent. It is far better when you can figure it out yourself. Yes, the elements of the quotient space of an equivalence relation are sets (the equivlance classes) forming a partition of the original space. For every element $a \in A, [a]$ is the equivalence class of $a$, which is $\{x \in A\mid x\sim a\}$. In all cases, $a \in [a]$ by reflexivity, if $x \in [a]$, then $a \in [x]$ by symmetry, and if $x \in [a]$ and $x \in [b]$, then $[a] = [b]$, by transitivity. So the intersection of any two classes for the same equivalence relation is empty. $\endgroup$ – Paul Sinclair Nov 14 '19 at 0:08

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