# Intersection of equivalence classes of two equivalence relations

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?

• 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. – William Elliot Nov 12 '19 at 3:57
• @WilliamElliot, thank you! I wouldn't see it. Lapsus calami... What about the rest? This was a part of a question. – Invisible Nov 12 '19 at 6:52
• It is incoherence. Statement iff statement iff set is nonsense. Statement implies statement iff statement iff statement is incoherent for lack of parenthesis. – William Elliot Nov 12 '19 at 10:57

K = $$\cap$$C is an equivalence relation and
for all x in S, [x]$$_K$$ = $$\cap$${ [x]$$_R$$ : R in C }.
• @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. – Invisible Nov 12 '19 at 14:56