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?
Best Answer
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 }.