Rank Function/Examples/Arbitrary Example 1

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Example of Rank Function

Let $X = \set {x, y, z}$.

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

$\RR = \set {\tuple {x, x}, \tuple {x, y}, \tuple {x, z}, \tuple {y, y}, \tuple {z, z} }$.

Let the mapping $\operatorname {rk}_0: X \to \N$ be defined as:

\(\ds \map {\operatorname {rk}_0} x\) \(=\) \(\ds 0\)
\(\ds \map {\operatorname {rk}_0} y\) \(=\) \(\ds 1\)
\(\ds \map {\operatorname {rk}_0} z\) \(=\) \(\ds 1\)

Then $\operatorname {rk}_0$ is a rank function for $\RR$.


Proof

There are two ordered pairs $\tuple {a, b}$ in $\RR$ such that $a \ne b$:

\(\text {(1)}: \quad\) \(\ds \tuple {x, y}: \, \) \(\ds \map {\operatorname {rk}_0} x\) \(=\) \(\ds 0\)
\(\ds \map {\operatorname {rk}_0} y\) \(=\) \(\ds 1\)
\(\text {(2)}: \quad\) \(\ds \tuple {x, z}: \, \) \(\ds \map {\operatorname {rk}_0} x\) \(=\) \(\ds 0\)
\(\ds \map {\operatorname {rk}_0} z\) \(=\) \(\ds 1\)

and the fact that $\operatorname {rk}_0$ is a rank function for $\RR$ is clear.

$\blacksquare$


Sources

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