Symmetric Closure of Ordering may not be Transitive

Theorem
Let $\left({S, \preceq}\right)$ 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
Proof by Counterexample:

Let $S = \left\{{a, b, c}\right\}$ where $a$, $b$, and $c$ are distinct.

Let:
 * ${\preceq} = \left\{{\left({a, a}\right), \left({b, b}\right), \left({c, c}\right), \left({a, c}\right), \left({b, c}\right)}\right\}$:

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 $\preceq$:


 * ${\preceq^\leftrightarrow} = {\preceq} \cup {\preceq}^{-1} = \left\{{\left({a, a}\right), \left({b, b}\right), \left({c, c}\right), \left({a, c}\right), \left({c, a}\right), \left({b, c}\right), \left({c, b}\right)}\right\}$

by inspection.

Now $\left({a, c}\right) \in {\preceq^\leftrightarrow}$ and $\left({c, b}\right) \in {\preceq^\leftrightarrow}$, but $\left({a, b}\right) \notin {\preceq^\leftrightarrow}$.

Thus $\preceq^\leftrightarrow$ is not transitive.