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Math Help - Distinct equivalent classes the null set

  1. #1
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    Distinct equivalent classes the null set

    Let R be an equivalence relation on a set S. Let E and F be two distinct equivalence classes of R. Prove that E and F = null set.

    Do i show that itys transitive im a bit stuck. Help much appreciated thanks.
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    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by adam_leeds View Post
    Let R be an equivalence relation on a set S. Let E and F be two distinct equivalence classes of R. Prove that E and F = null set.

    Do i show that itys transitive im a bit stuck. Help much appreciated thanks.
    Do you mean E \cap F = \emptyset?

    Assume there exists an element in E \cap F, call it x. Then you can apply transitivity as every element of E is equivalent to x, as is every element of F. e \sim x and x \sim f \Rightarrow e \sim f \Rightarrow E = F, a contradiction.
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    Quote Originally Posted by Swlabr View Post
    Do you mean E \cap F = \emptyset?

    Assume there exists an element in E \cap F, call it x. Then you can apply transitivity as every element of E is equivalent to x, as is every element of F. e \sim x and x \sim f \Rightarrow e \sim f \Rightarrow E = F, a contradiction.
    Yep thats what i meant thanks.
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    Quote Originally Posted by Swlabr View Post
    Do you mean E \cap F = \emptyset?

    Assume there exists an element in E \cap F, call it x. Then you can apply transitivity as every element of E is equivalent to x, as is every element of F. e \sim x and x \sim f \Rightarrow e \sim f \Rightarrow E = F, a contradiction.
    acually im a bit confused how is this a contradiction, it doesnt show its in the null set, just that e is in f?
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  5. #5
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    Quote Originally Posted by adam_leeds View Post
    acually im a bit confused how is this a contradiction, it doesnt show its in the null set, just that e is in f?
    You assumed that there is some element in E\cap F. As Swlabr showed, this implies E=F, however you took E,F to be distinct equivalence classes, ie. E \neq F - so this is a contradiction.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by adam_leeds View Post
    Let R be an equivalence relation on a set S. Let E and F be two distinct equivalence classes of R. Prove that E and F = null set.

    Do i show that itys transitive im a bit stuck. Help much appreciated thanks.
    More of a forward-knowledge looking back approach (since you need to do this problem to prove what I'm about to say), but a relation R on S induces a partition \pi of S where the blocks are the equivalence classes. If that is a definition in your book the answer follows immediately.
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