Past exam of the year 1 of the computer science course of the University of Oxford / 2017 / Discrete mathematics, continuous mathematics and probability / 3 / e

A preorder is reflexive and transitive: $xRx$ for every element, and $xRy,yRz$ imply $xRz$. A partial order additionally requires antisymmetry: $xRy$ and $yRx$ imply $x=y$. It need not compare every pair of distinct elements.

No. On $\{0,1,2\}$ take $R=\{(0,1),(1,2),(2,0)\}$. This is an antisymmetric relation because it contains no opposite pair of distinct elements. Its reflexive transitive closure contains every ordered pair: following the directed three-cycle reaches any vertex from any other. In particular it contains both $(0,1)$ and $(1,0)$, violating antisymmetry.

This illustrates the reachability preorder: collapsing mutually reachable vertices into strongly connected components yields a partial order, but retaining all original vertices need not.