Reflexive Relation on Set of Cardinality 2 is Transitive

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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.

$\blacksquare$