Transitive Closure (Relation Theory)/Examples/Arbitrary Example 2

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Example of Transitive Closure

Let $S = \set {1, 2, 3, 4, 5}$ be a set.

Let $\RR$ be the relation on $S$ defined as:

$\RR = \set {\tuple {1, 2}, \tuple {2, 3}, \tuple {3, 4}, \tuple {5, 4} }$

The transitive closure $\RR^+$ of $\RR$ is given by:

$\RR^+ = \set {\tuple {1, 2}, \tuple {1, 3}, \tuple {1, 4}, \tuple {2, 3}, \tuple {2, 4}, \tuple {3, 4}, \tuple {5, 4} }$


Proof

By Construction of Transitive Closure of Relation:

\(\ds \tuple {1, 2}\) \(\in\) \(\ds \RR^+\) Rule $(1)$
\(\ds \tuple {2, 3}\) \(\in\) \(\ds \RR^+\) Rule $(1)$
\(\ds \tuple {3, 4}\) \(\in\) \(\ds \RR^+\) Rule $(1)$
\(\ds \tuple {5, 4}\) \(\in\) \(\ds \RR^+\) Rule $(1)$

That completes the phase of the construction which uses Rule $(1)$.


Then:

\(\ds \tuple {1, 2}\) \(\in\) \(\ds \RR\) by hypothesis
\(\ds \tuple {2, 3}\) \(\in\) \(\ds \RR^+\) a priori
\(\ds \leadsto \ \ \) \(\ds \tuple {1, 3}\) \(\in\) \(\ds \RR^+\) Rule $(2)$


\(\ds \tuple {2, 3}\) \(\in\) \(\ds \RR\) by hypothesis
\(\ds \tuple {3, 4}\) \(\in\) \(\ds \RR^+\) a priori
\(\ds \leadsto \ \ \) \(\ds \tuple {2, 4}\) \(\in\) \(\ds \RR^+\) Rule $(2)$


\(\ds \tuple {1, 2}\) \(\in\) \(\ds \RR\) by hypothesis
\(\ds \tuple {2, 4}\) \(\in\) \(\ds \RR^+\) a priori
\(\ds \leadsto \ \ \) \(\ds \tuple {1, 4}\) \(\in\) \(\ds \RR^+\) Rule $(2)$

There are no further elements of $\RR$ for which there are corresponding elements of $\RR^*$ that satisfy Rule $(2)$.

Hence the result.

$\blacksquare$


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