Set of all symmetric and not reflexive relations

Question

Let A be A={1,2,3}, let K be the set of all symmetric and not reflexive relations of A. Is K $$ K=\{(\phi,\phi)\quad((\phi,\phi),(1,2),(2,1))\quad ((\phi,\phi),(1,3),(3,1))\quad ((\phi,\phi),(2,3),(3,2))\quad((\phi,\phi),(1,2),(2,1),(1,3),(3,1))\quad ((\phi,\phi),(1,2),(2,1),(2,3),(3,2))\quad ((\phi,\phi),(1,3),(3,1),(2,3),(3,2))\quad ((\phi,\phi),(1,2),(2,1),(1,3),(3,1),(2,3),(3,2))\} $$

???

Answer 1

If $R$ is a relation not reflexive, we cannot have $(1,1),(2,2),(3,3) \in R$ together but for example you may have $(1,1) \in R$ and $(2,2),(3,3) \notin R$. Or $(1,1),(2,2) \in R$ and $(3,3) \notin R$. So if you want to list them systematically, answer should also include the relations:

$$K=\big\{\{(1,1)\},\{(2,2)\},\{(3,3)\},\{(1,1),(2,2)\},\{(1,1),(3,3)\},\{(2,2),(3,3)\},\{(1,1),(1,2),(2,1)\},\{(1,1),(1,3),(3,1)\},\{(1,1),(2,3),(3,2)\},... \big\}$$

and none of the elements of $K$ should not include $(\emptyset,\emptyset)$ since $\emptyset$ is not an element but a representation of empty set.

Answer 2

Let $K$ be the set of relations $R$ on $A$ such that $R$ is symmetric, but not reflexive.

Then $R \in K$ if and only if $R=S\cup T$, where $S$ is a proper subset of $$\{(1,1),(2,2),(3,3)\}$$ and $T$ is the union of zero or more of the three sets $$\{(1,2),(2,1)\},\{(2,3),(3,2)\},\{(3,1),(1,3)\}$$

It follows that there are $2^3-1=7$ choices for $S$, and $2^3=8$ choices for $T$, so there are $(7)(8)=56$ possibilities for $R$.

Thus, $K$ has $56$ elements.