Reflexive, symmetrical but not transitive I am supposed to create a relation R on a set $X = \{a,b,c,d\}$ that is reflexive, symmetrical but not transitive. 
My attempt looks like this: $R = \{(a,a),(b,b),(c,c),(d,d),(a,b)(b,a),(b,c),(c,b),(c,d)(d,c)\}$
I believe my current solution works - but I have also been presented with a minimal solution, which would look like this:
$R = \{(a,a),(b,b),(c,c),(d,d),(a,b),(b,a),(b,c),(c,b)\}$
It reflexive since $(a,a),(b,b),(c,c),(d,d)\in R$.
I can also see that it is not transitive, since $(a,b) \in R$ and $(b,c) \in R $ but there is no $(a,c) \in R $. 
But is it symmetrical? 
The defintion of a symmetrical relationship is as follows:
$\forall x \forall y$ $xRy \implies yRx $  
I thought that since the element $d \in X$, I would have to include something like $(d,c)$ and $(c,d)$ in R, in order for it to actually be symmetrical. I interpret the $\forall x \forall y $ as all elements in X, am I missing something really obvious here? 
Is it perhaps that $(d,d)\in R$ and if we were to look at $x=d$ and $y=d$, we do in fact have a $xRy \implies yRx$ ? 
 A: To be symettric if $(d,c)$ is included than $(c,d)$ must also be included.  But there is absolutely no reason $(d,c)$ need to be included$.
To have a minimum relationship that is not transitive you need:
Wolog: $(a,b)$ and $(b,c)$ but not $(a,c)$.
To be reflexive you need.  $(a,a), (b,b), (c,c), (d,d)$.
Since you have $(a,b)$ and $(b,c)$ you need $(b,a)$ and $(c,b)$.  You also need $(a,a), (b,b), (c,c),(d,d)$ but those are "self-symmetric" so to speak and we already listed them.
so $(a,a)(b,b)(c,c)(a,b)(b,a),(a,c),(c,a),(d,d)$ is reflexive symmetric and not reflexive and minimal.
Now if we threw in any $(d,x)$ we would have to throw in $(x,d)$ but there is utterly no reason we have to throw in any $(d,x); d\ne x$.
Perhaps it would make things clear if we point out the ONLY reason we had to toss it $(a,b)$ in the first place was so that it couldn't be transitive.  If we don't have any $(x,y); x\ne y$ we can't have any $(x,y), (y,z)$ but not $(x,z)$.
If the problem was find a relationship that was reflexive and symmetric and we don't care whether it is or is not transitive, the minimal would be $\{(a,a),(b,b), (c,c),(d,d)\}$.
It's reflexive:  $(x,x)$ is included for all $x$.
It's symmetric:  If $(x,y)$ is included so is $(y,x)$.
But it is also transitive.
To not be transitive we need $(x,y)$ and $(y,z)$ without $(x,z)$.  (So $x\ne z$ as $(x,x)$ is included.  ANd $z \ne y$ as $(x,y)$ is included and $x \ne y$ as $(y,z)$ is included.)  But we don't need any more.
A: Consider $A$, $B$ and $C$ non-empty sets. Define the relation "i" as: $A$ is in the relation "i" with $B$ iff $A \cap B \neq \emptyset$. Thus, it is easy to construct examples with $A$i$B$, $B$i$C$ but there is no relation between $A$ and $C$.
A: I think you're misreading how the quantifiers interact with the implication in the definition of symmetry.
Requiring that $\forall x\forall y(xRy\Rightarrow yRx)$ is the same as requiring that all of these 16 instances must be true:
$$ \begin{array}{cccc}
aRa \Rightarrow aRa, & aRb \Rightarrow bRa, & aRc \Rightarrow cRa, & aRd \Rightarrow dRa, \\
bRa \Rightarrow aRb, & bRb \Rightarrow bRb, & bRc \Rightarrow cRb, & bRd \Rightarrow dRb, \\
cRa \Rightarrow aRc, & cRb \Rightarrow bRc, & cRc \Rightarrow cRc, & cRd \Rightarrow dRc, \\
dRa \Rightarrow aRd, & dRb \Rightarrow bRd, & dRc \Rightarrow cRd, & dRd \Rightarrow dRd \\
\end{array}$$
In both of the examples you present, some of these -- such as $aRb\Rightarrow bRa$ -- are true because both sides of the $\Rightarrow$ are true.  Others, such as $dRb\Rightarrow bRd$ are true because both sides are false, which is a perfectly good way for a $\Rightarrow$ to be true.
The difference between your example and the minimal example is just that $dRc\Rightarrow cRd$ and $cRd\Rightarrow dRc$ are true for one reason in your example and true for another reason in the minimal example. They're true all the same.
