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Equivalence relations

  1. Aug 22, 2010 #1
    Given is the set X. The set of functions from X to [0,1] we call Fun(X,[0,1]). On this set we consider the relation R. An ordered pair (f,g) belongs to R when [tex]f^{-1}(0)\setminus g^{-1}(0)[/tex] is a countable set.

    a) Prove that R is transitive.

    b) Is R an equivalence relation? Prove!

    c) Prove that [tex]R \cap R^{-1}[/tex] is an equivalence relation.

    2. Relevant equations
    Transitive means that if (f,g) and (g,h) belong to the relation, that also (f,h) belongs to it.

    Equivalence relation is a relation that is transitive, reflexive ([tex](f,f) \in R[/tex] and symmetric ([tex](f,g) \in R \Rightarrow (g,f) \in R[/tex].

    3. The attempt at a solution
    a) [tex]f^{-1}(0)\setminus g^{-1}(0)[/tex] is a countable set. So [tex]f{-1}(0)[/tex] is a countable set. This means that [tex]f^{-1}(0)\setminus h^{-1}(0)[/tex], because a subset of a countable set is also countable.

    Is this correct?

    b) Reflexivity is easy, because [tex]f^{-1}(0)\setminus f^{-1}(0)[/tex] is the empty set so that is obviously countable.
    How do I prove that it is symmetric?

    c) Is this just the subset of all the reflexive pairs?
    Last edited: Aug 22, 2010
  2. jcsd
  3. Aug 22, 2010 #2


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    You can't conclude [itex]f^{-1}(0)[/itex] is countable. For instance, suppose X is R, the set of real numbers, and f(x)=g(x)=0. Then [itex]f^{-1}(0)\setminus g^{-1}(0)[/itex] is the empty set, but [itex]f^{-1}(0)=R[/itex], which is not countable.

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