Reflexive Relation on Set of Cardinality 2 is Transitive
Theorem
Let $S$ be a set whose cardinality is equal to $2$:
- $\card S = 2$
Let $\odot \subseteq S \times S$ be a reflexive relation on $S$.
Then $\odot$ is also transitive.
Proof
Without loss of generality, let $S = \set {a, b}$.
Let $\odot$ be reflexive.
By definition of reflexive relation:
- $\Delta_S \subseteq \odot$
where $\Delta_S$ is the diagonal relation:
- $\Delta_S = \set {\tuple {x, x}: x \in S}$
That is:
- $\set {\tuple {a, a}, \tuple {b, b} } \subseteq \odot$
Suppose $\set {\tuple {a, a}, \tuple {b, b} } = \odot$.
Then by Diagonal Relation is Equivalence, $\odot$ is transitive.
Suppose there exists $\tuple {x, y} \in \odot$ such that $\tuple x \ne y$ and $\tuple {y, x} \notin \odot$.
Then:
- $x \odot x, x \odot y \implies x \odot y$
and:
- $x \odot y, y \odot y \implies x \odot y$
Now suppose there exists $\tuple {x, y} \in \odot$ such that $\tuple x \ne y$ and $\tuple {y, x} \in \odot$.
Then we have:
- $x \odot y, y \odot x \implies x \odot x$
and:
- $y \odot x, x \odot y \implies y \odot y$
which hold because $\odot$ is reflexive
Hence in all cases, a reflexive relation on $S$ is also transitive.
$\blacksquare$