Symmetric and Transitive therefore Reflexive

Fallacy
Let $$\mathcal R \subseteq S \times S$$ be a relation which is symmetric and transitive.

Then $$\mathcal R$$ is also always reflexive.

Consider $$x, y \in S$$.

Suppose $$x \mathcal R y$$.

Then as $$\mathcal R$$ is symmetric, it follows that $$y \mathcal R x$$.

As $$\mathcal R$$ is transitive, it follows that $$x \mathcal R x$$.

Therefore $$x \mathcal R x$$ and so $$\mathcal R$$ is reflexive.

Resolution
For $$\mathcal R$$ to be reflexive, it is necessary for $$x \mathcal R x$$ for all $$x \in S$$.

Unless it is the case that $$\forall x \in S: \exists y \in S: x \mathcal R y$$, it is not necessarily the case that also $$y \mathcal R x$$, and so the reasoning does not follow.

Take the set $$S = \left\{{0, 1}\right\}$$ and the relation $$\mathcal R \subseteq S \times S: x \mathcal R y \iff x = y = 1$$.

It can easily be seen that $$\mathcal R$$ is symmetric and transitive but not reflexive as $$\neg \left({0 \mathcal R 0}\right)$$.

Also see

 * Definition of equivalence relation, which is the usual motivator of this frequently-met fallacy.


 * Relation Symmetric Transitive and Reflexive which shows that the condition under which a symmetric and transitive relation is guaranteed to be reflexive.