Rank Function Property of Well-Founded Relation/Proof

Proof
$\RR$ is not a well-founded relation.

From Infinite Sequence Property of Well-Founded Relation, there exists an infinite sequence $\sequence {a_n}$ of elements of $S$ such that:


 * $\forall n \in \N: \paren {a_{n + 1} \mathrel \RR a_n} \text { and } \paren {a_{n + 1} \ne a_n}$

Let $A = \operatorname {rk} \sqbrk {\sequence {a_n} }$ denote the image of $\sequence {a_n}$ under $\operatorname {rk}$.

From Image is Subset of Codomain:
 * $A \subseteq T$

Let $m \in A$ be the smallest element of $A$.

Such a smallest element exists by the fact that $T$ is well-ordered by $\prec$.

Let $a_k \in S$ such that $\map {\operatorname {rk} } {a_k} = m$.

Then $a_{k + 1} \mathrel \RR a_k$.

But then:
 * $\map {\operatorname {rk} } {a_{k + 1} } \prec m$

by definition of $\operatorname {rk}$.

Both $\map {\operatorname {rk} } {a_{k + 1} } \in A$ and $m \in A$.

But $m$ is the smallest element of $A$ under $\prec$.

This contradicts $\map {\operatorname {rk} } {a_{k + 1} } \prec m$.

Hence by Proof by Contradiction our initial assumption that $\RR$ is not a well-founded relation was false.

Hence the result.