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19 November, 23:24

Ex. 4. Suppose that R1 and R2 are equivalence relations on the set S. Determine whether each of these combinations of R1 and R2 must be an equivalence relation. a) R1∪R2 b) R1∩R2 c) R1⊕R2

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  1. 19 November, 23:44
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    a) R1∪R2 may not be an equivalence relation.

    b) R1∩R2 is an equivalence relation.

    c) R1⊕R2 is not an equivalence relation.

    Step-by-step explanation:

    a)

    R1∪R2 may not be an equivalence relation.

    Counter-example

    S={1,2,3}

    R1={ (1,1), (2,2), (3,3), (1,2), (2,1) }

    R2={ (1,1), (2,2), (3,3), (1,3), (3,1) }

    R1 and R2 are equivalence relations in S. Let us see R1∪R2 is not.

    (2,1) ∈ R1∪R2 and (1,3) ∈ R1∪R2. If R1∪R2 were an equivalence relation in S it should be transitive, so (2,3) should belong to R1∪R2. But (2,3) ∉ R1∪R2 and R1∪R2 is not an equivalence relation

    b)

    R1∩R2 is an equivalence relation.

    Let us prove is reflexive:

    (a, a) ∈R1 for every a in S, (a, a) ∈R2 for every a in S, so

    (a, a) ∈R1∩R2

    is symmetric:

    If (a, b) ∈R1∩R2 then (a, b) ∈R1 and (a, b) ∈R2. Since R1 and R2 are equivalence relations, (b, a) ∈R1 and (b, a) ∈R2, hence (b, a) ∈R1∩R2

    is transitive:

    If (a, b) ∈R1∩R2 and (b, c) ∈R1∩R2 then (a, b) ∈R1 and (b, c) ∈R1 and (a, b) ∈R2 and (b, c) ∈R2, so (a, c) ∈R1 and (a, c) ∈R2, hence (a, c) ∈R1∩R2.

    c)

    R1⊕R2 is not an equivalence relation.

    Since R1⊕R2 = (R1∪R2) - (R1∩R2) the intersection of R1 and R2 is not in R1⊕R2, so the diagonal (the elements of the form (a, a) with a∈S) are not in R1⊕R2 hence it is not reflexive.
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