Equivalence Relation: Examining the Proof

[Pere Callahan]

Hi,

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 R\subset M\times M is a binary, symmetric, transitive relation. Then for any a,b \in M with a\sim _R b it follows by symmetry that b\sim _R a and thus by transitivity that a\sim _R a 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
Can someone give a hint?

 

[tiny-tim]

Hi Pere Callahan!

But why should a ~ anything?

 

[Pere Callahan]

Oh, okay. Thanks tiny-tim!

Do you think that R=\{\}\subset\{0\}^2 would be a valid counter example?

 

[HallsofIvy]

Oh, okay. Thanks tiny-tim!

Do you think that R=\{\}\subset\{0\}^2 would be a valid counter example?

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.