Symmetric and Transitive therefore Reflexive
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Fallacy
Let $\RR \subseteq S \times S$ be a relation which is symmetric and transitive.
Then $\RR$ is also always reflexive.
Consider $x, y \in S$.
Suppose $x \mathrel \RR y$.
Then as $\RR$ is symmetric, it follows that $y \mathrel \RR x$.
As $\RR$ is transitive, it follows that $x \mathrel \RR x$.
Therefore $x \mathrel \RR x$ and so $\RR$ is reflexive.
Resolution
For $\RR$ to be reflexive, it is necessary for $x \mathrel \RR x$ for all $x \in S$.
Unless it is the case that $\forall x \in S: \exists y \in S: x \mathrel \RR y$, it is not necessarily the case that also $y \mathrel \RR x$, and so the reasoning does not follow.
Take the set $S = \set {0, 1}$ and the relation:
- $\RR \subseteq S \times S: x \mathrel \RR y \iff x = y = 1$
It is seen that:
- $\RR$ is symmetric
- $\RR$ is transitive
but:
- $\RR$ is not reflexive, as $\neg \paren {0 \mathrel \RR 0}$.
$\blacksquare$
Also see
- Definition: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
- 1964: W.E. Deskins: Abstract Algebra ... (previous) ... (next): Exercise $1.2: 6$
- 1966: Richard A. Dean: Elements of Abstract Algebra ... (previous) ... (next): $\S 0.3$: Example $5$
- 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{I}$: Sets and Functions: Exercise $\text{F}$
- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 6$. Indexed families; partitions; equivalence relations: Exercise $6$
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): Chapter $1$: Sets and mappings: $\S 1.4$: Equivalence relations: Exercise $4$
- 2000: James R. Munkres: Topology (2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 3$: Relations: Exercise $3.3$