If $A$ is a set $\{2,4,6,8\}$, and we are asked to give a relation on $A$ that is:
symmetric, antisymmetric, but not reflexive, is this possible?
If we were to say $\{(2,2),(4,4)\}$, it would indeed be symmetric and antisymmetric, but it would also be reflexive.
If we were to say $\{(2,2),(4,4),(6,8)\}$, this would not be symmetric.
I'm thinking that this isn't possible, but would like to know others' thoughts.
On
Ah, but $\{ (2,2), (4,4) \}$ isn't reflexive on the set $\{2,4,6,8\}$ because, for example, $(6,6)$ is not in the relation.
On
Symmetric means that for all $a\neq b$, $R(a,b)\equiv R(b,a)$. Antisymmetric means that for all $a\neq b$, $R(a,b)\rightarrow \neg R(b,a)$. So if a relation is both symmetric and antisymmetric, you necessarily have $R(a,b)\rightarrow \neg R(a,b)$ for all $a\neq b$, and hence $R(a,b)$ is false for all $a\neq b$.
This doesn't tell you anything about whether $R$ is reflexive or not. You can choose the value of $R(a,a)$ to be true or false for each $a$ independently; if you choose true in every case, $R$ is reflexive; otherwise it isn't.
Recall that being symmetric and anti-symmetric are internal properties of the relation. They can be defined regardless to any external set (such as $A$). These properties merely say "If the ordered pair ... is in the relation, then there is another ordered pair ..." or so.
On the other hand, reflexivity is an external property. It must refer to an external set. We say that $R$ is a reflexive relation on $A$ if every $a\in A$ satisfies $(a,a)\in R$. So for example $\varnothing$ is a reflexive relation on $\varnothing$, but not on $\{a\}$.