Symmetric Closure of Ordering may not be Transitive

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

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

Let $\preceq^\leftrightarrow$ be the symmetric closure of $\preceq$.

Then $\preceq$ is an ordering but $\preceq^\leftrightarrow$ is not transitive.

Proof
$\preceq$ 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.

$\preceq^\leftrightarrow$, the symmetric closure of $\preceq$, is defined by:
 * ${\preceq^\leftrightarrow} = {\preceq} \cup {\preceq}^{-1} = \left\{{(a,a),(b,b),(c,c),(a,c),(c,a),(b,c),(c,b)}\right\}$

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

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