A relation on the set of real numbers which is only reflexive.

Question

Can we construct a relation R on the set of real numbers such that it is only reflexive & neither symmetric nor transitive?

Answer 1

I claim that almost any attempt to build such a relation will succeed. Take any reflexive relation, select two distinct elements and use them to break symmetry, and select three other distinct elements and use them to break transitivity.

Answer 2

A binary relation on the real numbers is any subset of ${\Bbb R}{\times}{\Bbb R}$.

Let $R$ be any reflexive, symmetric, and transitive relation on ${\Bbb R}$. Select two arbitrary but distinct elements $a$ and $b$ from ${\Bbb R}$ such that $(a,b)\in R$ and construct a new relation $R_1$ as follows:

$$ R_1 = R \setminus\{(a,b)\}$$

Now $R_1$ is reflexive because $R$ is.

$R_1$ is not symmetric since $(b,a)\in R_1$ (because $(b,a)\in R$ being symmetric) but $(a,b) \notin R_1$.

To show that $R_1$ is not transitive, take any element $c\neq a,b\in {\Bbb R}$ such that $(a,c)\in R$ and $(c,b)\in R$.

From this construction, we see that it's necessary to require that the domain of $R$ has at least cardinality three otherwise we wouldn't be able to pick the $a, b, c$ needed for the non-symmetry and non-transitivity of $R_1$.