Question

Q. show that the intersection of two equivalence relations in a set is again an equivalence relation. please answer correctly​

Answer 1

Answer:

If not an equivalence relation, then R∩S fails to be reflexive and/or fails to be symmetric, and/or fails to be transitive. If you can work towards a contradiction (that this assumption must contradict the fact that both R and S are equivalence relations), then you are done.

Answer 2

Suppose R & S are both equivalence relations on a set A. In what follows we will show that R & S both being equivalence relations on the set A implies R∩S is also an equivalence relation.

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▶It is immediately apparent that since R&S are equivalence relation 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 & S are symmetric so it most 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..

({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.

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