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
, 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.