I pretty much know how to get the ordered pairs by doing the arrow graph method since the matrix method is much more complex.
let R be:
R = { (a,b), (b,a), (a,c), (c,d), (c,e), (e,c) }
(I am missing a black arrow from e to c btw)
Since we know that we can get from A to D by going through C, we add another arrow (arrows in red are the ones added because of set rules). Same for B. Repeating this process without going in the opposite orientation of the arrows, we get the following (I added the missing black arrow too btw)
Therefore, we start writing out the pairs.
R* = { (a,b), (a,c), (a,d), (a,e), (b,a), (b,c), (b,d), (b,e), (c,d), (c,e), (e,c), (e,d) }
This is what I dont understand. My professor while in class had red arrows pointing to the relation elements (for example, a pointing to itself) for a,b,c and e. Having this, it would add to R* the following pairs: (a,a), (b,b), (c,c) and (e,e).
My question is why? what determines that an element should be pointing an arrow to itself? I know that this is called reflexive, but I have no idea what should I consider in order to add the red arrows pointing to the element itself.
Thank you for your help!
As transitive closure of relation $R$ the relation $R^*$ is transitive and satisfies $R\subseteq R^*$. In fact it is the 'smallest' transitive relation that contains $R$. If we have $uRv$ and $vRw$ then also $uR^*v$ and $vR^*w$. The transitivity of $R^*$ then tells us that $uR^*w$.
Applying that here on cases like $u=a,v=b,w=c$ (or $u=e,v=c,w=e$) leads to the conclusion that $aR^*a$ (or $eR^*e$).