I'm familiar with Number of relations on A that are reflexive and symmetric but not transitive but I'm not clear on how to close out the problem.
Let $S$ be a set with $5$ elements. How many relations on $S$ are reflexive and symmetric but not transitive?
I totally agree that relations that are all three are equivalence relations and are enumerated by the Bell number $B(5)$. So my thinking is that I'll add up all relations that are reflexive, $2^{20}$, and then add up all the relations that are symmetric, $2^{\binom{5}{2}}$, and then subtract $B(5)$. My problem with this though is that amidst the $2^{20}$ relations that are reflexive, some are ALSO symmetric, and I'm having a lot of trouble figuring out what else I need to subtract by.
Relation $R$ over $S$ is reflexive and symmetric when:$$\forall x{\in}S~.(\langle x,x\rangle{\in}R\wedge\forall y{\in} S~. (y\succ x\wedge \langle x,y\rangle{\in} R\to\langle y,x\rangle{\in} R))$$
That is to say, if we represent a relation as a $5\times 5$ grid with a check in every cell where the ordinates are in the relation, then the above will be true when every diagonal cell is checked, and a selection from the 10 cells in the upper triangle are check, as are the corresponding cells in the lower triangle.
So just count the ways to do this.