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 $\neg \left({0 \mathcal R 0}\right)$.

$\blacksquare$

Also see

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

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

Sources

• W.E. Deskins: Abstract Algebra (1964): $\S 1.2$: Exercise $1.2: 6$
• Richard A. Dean: Elements of Abstract Algebra (1966): $\S 0.3$: Example $5$
• George McCarty: Topology: An Introduction with Application to Topological Groups (1967): $\text{I}$: Exercise $\text{F}$
• T.S. Blyth: Set Theory and Abstract Algebra (1975): $\S 6$: Exercises $5, \ 6$