Transitive Closure of Reflexive Symmetric Relation is Equivalence

Theorem
Let $S$ be a set.

Let $\mathcal R$ be a symmetric and reflexive relation on $S$.

Then the transitive closure of $\mathcal R$ is an equivalence relation.

Proof
Let $\sim$ be the transitive closure of $\mathcal R$.

Checking in turn each of the criteria for equivalence:

Reflexivity
By Transitive Closure of Reflexive Relation is Reflexive:
 * $\sim$ is reflexive.

Symmetry
By Transitive Closure of Symmetric Relation is Symmetric:
 * $\sim$ is symmetric.

Transitivity
By the definition of transitive closure:
 * $\sim$ is transitive.

$\sim$ has been shown to be reflexive, symmetric and transitive.

Hence by definition it is an equivalence relation.