Relations which are not reflexive but are symmetric and antisymmetric at the same time 
How many relations are there on a finite n-element set, such that they are not reflexive but are symmetric and antisymmetric at the same time?

In my opinion if a relation is symmetric and antisymmetric at the same time, it should be reflexive. But I saw this question and I am stuck.
 A: After reading your comment, I see where we have a problem. Unfortunately, this is a very common misconception. But in fact, symmetric and antisymmetric together do NOT imply reflexive. The reason is that both symmetric and antisymmetric properties are "if-then" statements. For example, the symmetric property for a relation $R$ on a set $S$ states:

Symmetric: If $(x,y)\in R$, then $(y,x)\in R$.

Note that it does not force $(x,y)$ to be in $R$; only if it happens to be in $R$, then the other pair should be included too. Similarly, the antisymmetric property is:

Antisymmetric: If $(x,y)\in R$ and $(y,x)\in R$, then $(x,x)\in R$.

Again, it does not force $(x,x)$ to be in $R$; only if some conditions are met, then it should be included in $R$.
But the reflexive property requires that $\color{red}{(x,x)\in R \text{ for all } x\in S}$. The "for all" clause distinguishes the reflexive property from antisymmetric, where it is not required.
By the way, there's an equivalent statement of the antisymmetric property, in a way that doesn't make it look confusingly similar to the reflexive property: for any $x,y\in S$, $x\neq y$, we can't have both $(x,y)$ and $(y,x)$ in $R$. Having neither one or just one of them is fine, but not both.
Back to your question. Not a full solution, but a hint in the form of an example. Let $S=\{a,b\}$, and let $R$ be a relation on $S$ given by $R=\{(a,a)\}$. Note that it satisfies both the symmetric and antisymmetric properties, but it's not reflexive, because a reflexive relation on $S$ has to include both $(a,a)$ and $(b,b)$. Another (kinda extreme) example: even the empty relation $R=\varnothing$ satisfies the requirements of this question!
