Relation on Set of Cardinality 1 is Symmetric and Transitive
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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 $S$.
From Null Relation is Antireflexive, Symmetric and Transitive, $\odot$ is antireflexive, symmetric and transitive.
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.
$\blacksquare$