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This is a textbook problem, and I'm trying to understand how it can be proved. I feel like I can come up with a counterexample, but assume I must be wrong somewhere in my understanding.

For sake of counterexample, assume A = {1,2}, R={(1,1)}, S={(2,2)}, then R and S are equivalence relations (as they are both reflexive, symmetric and transitive). However, the intersection of R and S is the empty set. Since an equivalence relation cannot be the empty set, then $R\cap S$ is not an equivalence relation.

So... what is my mistake here?

Alex Butterfield
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    List the axioms for equivalence relations. Then verify them one by one. – Jacob Wakem Nov 03 '16 at 05:26
  • How can I do that when there are no specifics for the set? Also, why is the example I give not an adequate counterexample? (I'm assuming its not) – Alex Butterfield Nov 03 '16 at 07:44
  • By logical argument on x (R and S) y implies xRy and xSy implies (some propertry of equivalence relations for R) and (some property of equivalence relations for S) implies (some property of equivalence relations for (R and S) ) . – Jacob Wakem Nov 03 '16 at 16:50

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$R$ and $S$ are not equivalence relations. $(2,2)\not\in R$ and $(1,1)\not\in S$.

  • I'm not sure I agree. (yet) The definition of an equivalence relation doesn't say that they have to be the same set, as you are implying. – Alex Butterfield Nov 03 '16 at 06:45
  • DEFINITION A relation R on a set A is an equivalence relation on A if R is reflexive on A, symmetric, and transitive. – Alex Butterfield Nov 03 '16 at 06:45
  • @AlexButterfield They don't have to be the same set. But they do have to be reflexive. Your R and S aren't. –  Nov 03 '16 at 18:47
  • Thanks guys. I didn't realize that to be reflexive on a set A then all elements from set A must be in the relation in the form (x,x). I thought it was just, if some (x,y) is in R then (x,x) and (y,y) must be in R. – Alex Butterfield Nov 08 '16 at 09:07