Reflexive, symmetric and transitive closure of the following graph

Question

First of all, I have to understand which relation this graph represents:

Since there are 4 different elements, suppose there's the set $A = \{ a, b, c, d\}$

The relation drawn above, I think, can be described as follows on $A^2$:

$R_A = \{ (a, b), (b, a), (a, d), (b, d)\}$

I have first a more formal question: what would be the set build for this relation?

Now, since I am a newbie, I will just try to give the reflexive, symmetric and transite closure for the relation or to discuss the problems that I have:

1. Reflexive closure:

   I think it could be describe just adding the identity pairs:

   $R_R = \{ (a, b), (b, a), (a, d), (b, d), (a, a), (b, b), (c, c), (d, d)\}$

2. Symmetric closure:

   Here, I know I have to include all the "reversed" pairs of the presents ones, so my solution would be:

   $S_R = \{ (a, b), (b, a), (a, d), (d, a), (b, d), (d, b)\}$

3. Transitive closure:

   I think I have to include $(a, a)$ because we have $(a, b)$ and $(b, a)$. I have to include $(a, d)$, because we also have $(b, d)$. I also have to include $(b, d)$, because I have both $(b, a)$ and $(a, d)$. So the final closure would be:

   $T_R = \{ (a, b), (b, a), (a, d), (b, d), (a, a), (b, b)\}$

If you could try to correct me, if I am wrong, it would be great :)