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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$

$P,Q,R$ are classes of equivalence (respectively) $$Q\in[a]_1,\; R\in[a]_2$$

By definition, an equivalence relation is reflexive, symmetric and transitive.

reflexive property: $$\forall x\in A\; x\sim x\;\implies x\in[x]$$ symmetric property: $$\forall x,y\in A\;x\sim y\;\wedge\;y\sim x\;\;[x]=[y]$$

transitive property:

$$\forall x,y,z\in A\; x\sim y\;\wedge\; y\sim z\implies x\sim z$$

It is analogous for the $\thickapprox$ relation. Therefore, the conjunction holds the properties of both $\sim\;\&\thickapprox$. With: $$[a]:=\{x\in A: a\sim x\;\wedge\;a\thickapprox x\}\iff\{(x\in A:\;\;a\sim x)\;\wedge\;(x\in A:\;a\thickapprox x)\}\; $$ $$\iff\{\;[a]_1\;\cap\;[a]_2\;\}$$ $A_{/\triangle}=\{[a] : a\in A\}=\{\;[a]_1\;\cap\;[a]_2\;\}\implies P\in A_{/\triangle}\iff\;P\in ([a]_1\cap\;[a]_2)\;$ $\iff \exists \;Q\in A_{/\sim}\;\wedge\;R\in A_{/\thickapprox}\;$so that $\;P=Q\cap R$

Is this legitimate?

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  • $\begingroup$ 2 is false for left side has P as an element of A and the right side has P as an intersection of two subsets of A. $\endgroup$ – William Elliot Nov 12 '19 at 3:57
  • $\begingroup$ @WilliamElliot, thank you! I wouldn't see it. Lapsus calami... What about the rest? This was a part of a question. $\endgroup$ – Praskovya2.718281828 Nov 12 '19 at 6:52
  • $\begingroup$ It is incoherence. Statement iff statement iff set is nonsense. Statement implies statement iff statement iff statement is incoherent for lack of parenthesis. $\endgroup$ – William Elliot Nov 12 '19 at 10:57
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Generalizion.
If C is a collection of equivalence relations, all for the same set S, then
K = $\cap$C is an equivalence relation and
for all x in S, [x]$_K$ = $\cap${ [x]$_R$ : R in C }.

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  • $\begingroup$ So, this way, I look at equivalence relations as elements of the set C, find the intersection of all elements - equivalences, which is also an equivalence relation? K is the number of those intersections for each equivalence relation, and R is the number of all equivalences in C ? Thank you very much for corrections. $\endgroup$ – Praskovya2.718281828 Nov 12 '19 at 14:16
  • $\begingroup$ @Praskovya2.718281828. K and R are not numbers. $\endgroup$ – William Elliot Nov 12 '19 at 14:46
  • $\begingroup$ @WilliamElliotMy question was wrong, I thought $K$ in the index meant the number of intersections, which is obvious nonsense, even because of the intersection in the singular. Now, when it's a bit clearer, how can I prove that the intersection of a collection of equivalence relations over the same set is also an equivalence relation? I think this is what the initial question should've been. $\endgroup$ – Praskovya2.718281828 Nov 12 '19 at 14:56

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