Isn't reflexivity and symmetry implied in equivalence relations? It looks like for all "nice" sets, the set $S\times S$ will have symmetry and reflexivity by default. The tough part is usually showing transitivity. However, are there any non-empty sets such that $S\times S$ is not symmetric and reflexive? 
 A: The set $S\times S$ is always an equivalence relation, and is reflexive, symmetric, and transitive: it is the relation that says that every element of $S$ is related to every element of $S$.  

To answer your question from the comments, suppose we have some set $S$, and we have a relation on $S$, say $\sim$.  We want to represent $\sim$ as a subset $R_\sim\subset S\times S$, and your question is which subset $R_\sim$ corresponds to the relation $\sim$.  The correspondence is very straightforward: We have $a \sim b$ if and only if $\langle a,b\rangle \in R_\sim$.
For example, let $R_\le$ be the subset of $\Bbb Z\times \Bbb Z$ that represents the $\le$ relation.  Then $\langle 1,2\rangle$ and $\langle 1,17\rangle$ and $\langle 9,9\rangle$ are elements of $R_\le$, because $1\le2$ and $1\le 17$ and $9\le 9$, but $\langle 3,1\rangle$ is not because $3\not\le 1$.

When $R = S\times S$, we have that $\langle a,b\rangle\in R$ for all $a$ and $b$, and so the relation represented by this $R$ has $a\sim b$ for all $a$ and $b$—that is, $a$ and $b$ are always related. It should be easy to prove that this relation is reflexive, symmetric, and transitive.  Similarly, when $R = \emptyset$, we get the relation where $a$ and $b$ are never related, which is symmetric and transitive, but not reflexive.
