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Equivalence relation

  1. Nov 10, 2008 #1

    When I was just walking through the hallway of my department I found an exercise sheet asking the student to examine the following proof.

    Assume [itex]R\subset M\times M[/itex] is a binary, symmetric, transitive relation. Then for any [itex]a,b \in M[/itex] with [itex]a\sim _R b[/itex] it follows by symmetry that [itex]b\sim _R a[/itex] and thus by transitivity that [itex]a\sim _R a[/itex] i.e. R is also reflexive and therefore an equivalence relation.

    The exercise then asks to find the flaw in this argument (and give a counter example). To me the argument makes perfect sense...I am really ashamed, after all this is for first year students:smile:
    Can someone give a hint?
  2. jcsd
  3. Nov 10, 2008 #2


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    Hi Pere Callahan! :smile:

    But why should a ~ anything?
  4. Nov 10, 2008 #3
    Oh, okay. Thanks tiny-tim!

    Do you think that [itex]R=\{\}\subset\{0\}^2[/itex] would be a valid counter example?
  5. Nov 10, 2008 #4


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    Yes, the relation defined by the empty set is trivially symmetric and transitive but not reflexive. Here's a less trivial example: let A= {a, b, c} and "~" be the relation {(b,b), (b,c), (c,b), (c,c)}. That is both symmetric and transitive but not reflexive.
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