# Relation on Set of Cardinality 1 is Symmetric and Transitive

## Theorem

Let $S$ be a set whose cardinality is equal to $1$:

$\card S = 1$

Let $\odot \subseteq S \times S$ be a relation on $S$.

Then $\odot$ is both symmetric and transitive.

## Proof

Without loss of generality, let $S = \set a$.

There are $2$ relations on $S$:

$(1): \quad \odot := \O$, which is the null relation on $S$.

Thus in this case $\odot$ is both symmetric and transitive.

$(2): \quad \odot := \set {\tuple {a, a} }$, which is the trivial relation on $S$.

From Trivial Relation is Equivalence, $\odot$ is reflexive, symmetric and transitive.

Thus in this case also, $\odot$ is both symmetric and transitive.

$\blacksquare$