Symmetric Closure of Ordering may not be Transitive
Theorem
Let $\struct {S, \preceq}$ be an ordered set.
Let $\preceq^\leftrightarrow$ be the symmetric closure of $\preceq$.
Then it is not necessarily the case that $\preceq^\leftrightarrow$ is transitive.
Proof
Let $S = \set {a, b, c}$ where $a$, $b$, and $c$ are distinct.
Let:
- ${\preceq} = \set {\tuple {a, a}, \tuple {b, b}, \tuple {c, c}, \tuple {a, c}, \tuple {b, c} }$:
Then $\preceq$ is an ordering, but $\preceq^\leftrightarrow$ is not transitive, as follows:
$\preceq$ is reflexive because it contains the diagonal relation on $S$.
That $\preceq$ is transitive and antisymmetric can be verified by inspecting all ordered pairs of its elements.
Thus $\preceq$ is an ordering.
Now consider $\preceq^\leftrightarrow$, the symmetric closure of $\preceq$:
- ${\preceq^\leftrightarrow} = {\preceq} \cup {\preceq}^{-1} = \set{\tuple {a, a}, \tuple {b, b}, \tuple {c, c}, \tuple {a, c}, \tuple {c, a}, \tuple {b, c}, \tuple {c, b} }$
by inspection.
Now $\tuple {a, c} \in {\preceq^\leftrightarrow}$ and $\tuple {c, b} \in {\preceq^\leftrightarrow}$, but $\tuple {a, b} \notin {\preceq^\leftrightarrow}$.
Thus $\preceq^\leftrightarrow$ is not transitive.
$\blacksquare$