Transitive closure of a reflexive antisymmetric relation

Question

Prove that the transitive closure $S$ of a reflexive antisymmetric relation $R$ is a partial order.

Answer 1

The conjecture is false.

Let $A = \{1,2,3\}$ and let $R = \{(1,1),(2,2),(3,3),(1,2),(2,3),(3,1)\}$ which is reflexive and antisymmetric.

Its transitive closure, S = $\{(1,1),(2,2),(3,3),(1,2),(2,3),(3,1),(1,3),(2,1), (3,2)\}$, however is not antisymmetric since $(2,1)\in S$ and $(1,2)\in S$ but $1\neq 2$ and thus can't be a partial order.