Show that interaction of two equivalence relation on a set is again an equivalence relation but their Union may not be true
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Answer:
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Step-by-step explanation:
Proof. Suppose R and S are both equivalence relations on a set A. In what follows we will show that R and S both being equivalence relations on the set A implies that R∩S is also an equivalence relation.
It is immediately apparent that since R and S are equivalence relations then
x∈A⇒(xRx)∧(xSx)⇒(⟨x,x⟩∈R)∧(⟨x,x⟩∈S)⇒⟨x,x⟩∈R∩S.
Thus x∈A⇒x(R∩S)x therefore R∩S is reflexive.
Now suppose ⟨x,y⟩∈R∩S then (⟨x,y⟩∈R)∧(⟨x,y⟩∈S) by definition of the intersection of two sets. But both R and S are symmetric so it must be that (⟨y,x⟩∈R)∧(⟨y,x⟩∈S) which implies ⟨y,x⟩∈R∩S. Thus we have shown x(R∩S)y⇒y(R∩S)x therefore R∩S is symmetric.
Finally, consider (⟨x,y⟩∈R∩S)∧(⟨y,z⟩∈R∩S). Then since (⟨x,y⟩∈R)∧⟨(y,z⟩∈R)⇒⟨x,z⟩∈R, since R is transitive, and since (⟨x,y⟩∈S)∧⟨(y,z⟩∈S)⇒⟨x,z⟩∈S, since S is transitive, therefore ⟨x,z⟩∈R and ⟨x,z⟩∈S so by definition of the intersection of two sets ⟨x,z⟩∈R∩S. Thus we have shown (⟨x,y⟩∈R∩S)∧(⟨y,z⟩∈R∩S)⇒⟨x,z⟩∈R∩S.
It follows that R∩S is an equivalence relation since it is reflexive, symmetric and transitive. □