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.