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Equivalence Relation, prove dom(R) = range(R) = X

  1. Oct 8, 2011 #1
    1. The problem statement, all variables and given/known data
    Given:
    R is an equivalence relation over a nonempty set X

    Prove:
    dom(R) = X
    and range(R) = X


    2. Relevant equations


    3. The attempt at a solution
    I have the following thoughts:
    About the given:
    Since R is an equivalence relation over X by hypothesis, R satisfies:
    Reflexivity: <x,x> belongs to R
    Symmetry: <x,y> belongs to R, and <y,x> belongs to R
    Transitivity: <x,y> belongs to R, <y,z> belongs to R, and <x,z> belongs to R

    with x, y, z E X

    About the conclusion:
    Base on definition of domain and range of a relation R over a set X, I have:
    dom(R) = {x E X : there exists y belongs to Y such that <x,y> E R}
    range(R) = {y E Y : there exists x belongs to X such that <x,y> E R}


    What I'm confused is that I don't know how to connect my ideas together. The properties that R satisfies is with x, y, and z E X. And R is a subset of X x X. There is no Y whatsoever. So what should I do (or say) next to come to the conclusion?


    Thank you for your help.
     
  2. jcsd
  3. Oct 8, 2011 #2

    Dick

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    Reflexive is ALL you need. <x,x> is in R for ALL x in X, yes?
     
  4. Oct 9, 2011 #3
    I'm sorry but can you explain a little bit more?

    I think that since for all x in X, <x,x> is always in R holds, then according to the conditions of the domain, there indeed exists x E X such that there also exists a Y (namely, y = x) that makes <x,y> E R. But how about the range?

    thank you
     
  5. Oct 9, 2011 #4

    Dick

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    If <a,b> is in R, then 'a' is in the domain and 'b' is in the range, right?
     
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