Reflexive Transitive Closure/Examples/Arbitrary Example 2

Example of Reflexive 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 reflexive transitive closure $\RR^*$ of $\RR$ is given by:

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

Proof

From Transitive Closure of Relation: Arbitrary Example $2$, the transitive closure $\RR^+$ of $\RR$ is given by:

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

By definition of reflexive transitive closure:

$\RR^* = \paren {\RR^+}^=$

where $\paren {\RR^+}^=$ denotes the reflexive closure of $\RR^+$.

The diagonal relation on $S$ is given by:

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

The result then follows by definition of reflexive closure:

$\paren {\RR^+}^= := \RR^+ \cup \Delta_S$

$\blacksquare$

Sources

• 1979: John E. Hopcroft and Jeffrey D. Ullman: Introduction to Automata Theory, Languages, and Computation ... (previous) ... (next): Chapter $1$: Preliminaries: Exercises: $1.7$