Reflexive Closure of Transitive Relation is Transitive
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Theorem
Let $S$ be a set.
Let $\RR$ be a transitive relation.
Let $\RR^=$ be the reflexive closure of $\RR$.
Then $\RR^=$ is also transitive.
Proof
Let $a, b, c \in S$.
Suppose that $a \mathrel {\RR^=} b$ and $b \mathrel {\RR^=} c$.
If $a = b$, then since $b \mathrel {\RR^=} c$, also $a \mathrel {\RR^=} c$.
If $b = c$, then since $a \mathrel {\RR^=} b$, also $a \mathrel {\RR^=} c$.
The only case that remains is that $a \ne b$ and $b \ne c$.
Then by the definition of $\RR^=$, $a \mathrel \RR b$ and $b \mathrel \RR c$.
Since $\RR$ is transitive, it follows that:
- $a \mathrel \RR c$
and hence also $a \mathrel {\RR^=} c$.
Thus $\RR^=$ is transitive.
$\blacksquare$
Also see
Sources
- 1971: Robert H. Kasriel: Undergraduate Topology ... (previous) ... (next): $\S 1.19$: Some Important Properties of Relations: Exercise $2$