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
- 1996: Winfried Just and Martin Weese: Discovering Modern Set Theory. I: The Basics ... (previous) ... (next): Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $16$