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Symmetric/Antisymmetric Relations, Set Theory Problem

  1. Apr 3, 2007 #1
    1. The problem statement, all variables and given/known data
    Prove that if R is a symmetric relation on A, and Dom(R) = A, then R = the identity relation.

    2. The attempt at a solution
    My problem is... I don't believe the claim. At all. If A = {1, 2, 3} and R = {(1, 2), (2, 1), (3, 1), (1, 3)}, that satisfies the antecedent, and isn't the identity relation. Am I missing something? I can't exactly prove something I don't believe. Thanks for any help or explanations you can provide.

    p.s. This book has been known to have typos eeeeeverywhere. Suggestions as to what they really meant (antisymmetric? that wouldn't even work I don't think) are appreciated as well.
  2. jcsd
  3. Apr 3, 2007 #2


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    What exactly do you mean by Dom(R)=A?
  4. Apr 3, 2007 #3
    Domain of R is the entire set A.
  5. Apr 3, 2007 #4


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    I don't know what they could mean. Take any relation that satisfies the premise (such as yours, or the identity relation) and add pairs of relations, keeping it symmetric, and it still satisfies the premise. It seems like the Dom(R) part is crucial (since the symmetric part is pretty straightforward), so are you sure you've interpreted this right? You probably have, I'm just not familiar with that notation, and I don't know what else to suggest.
  6. Apr 3, 2007 #5
    I've decided that it'd be easier to prove that if R is symmetric and antisymmetric R = identity relation (it's the only thing I can think of). That should hold true shouldn't it?
  7. Apr 3, 2007 #6


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    Yea, and the domain still must be all of A.
  8. Apr 4, 2007 #7

    matt grime

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    Gah, whoever uses subsets to define relations should be taken outside and have their maths qualifications thoroughly slapped.

    Any equivalence relation is symmetric, and has domain A, surely. So the claim can't be true.
    If we have a relation that is reflexive (a~a) but satisfies a does not ~ any other b, then yes, only the identity satisfies that.
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