Question

I had a discussion question that follows:

What is wrong with the following argument? Note the argument supposedly shows that any symmetric and transitive relation R on A must also be reflexive.

Let R be a relation in A × A that is symmetric and transitive. Using symmetry, if (x,y) ∈ R then (y,x) ∈ R. Hence, both (x,y) and (y,x) are in R. Since (x,y) and (y,x) ∈ R, by transitivity, we have (x,x) ∈ R. Therefore, R is reflexive

An answer is:

The arguments proof states that (x,y) and (y,x) both exist in set R. However the argument proof is flawed.

Given: X={m,n,o} therefor we can assume R={ (m,m), (m,n), (n,m), (n,n) }. R is symmetric and transitive. The issue with the proof is that the element "o" does not exist in set R. For it truly to be reflexive it must contain the elements { (o,o), (o.m), (o,n), (m,o), (n,o) }.

My question is why are R={ (m,m), (m,n), (n,m), (n,n) } the only answers. why is (o,o) or any pairs with o not in the set. How do you know what pairs are in the set.

Answer 1