Strictly Increasing Mapping on Well-Ordered Class
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Theorem
Let $\struct {S, \prec}$ be a strictly well-ordered class.
Let $\struct {T, <}$ be a strictly ordered class.
Let $f$ be a mapping from $S$ to $T$.
For each $i \in S$ such that $i$ is not maximal in $S$, let:
- $\map f i < \map f {\map \Succ i}$
where $\map \Succ i$ is the immediate successor element of $i$.
Let:
- $\forall i, j \in S: i \preceq j \implies \map f i \le \map f j$
Then for each $i, j \in S$ such that $i \prec j$:
- $\map f i < \map f j$
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Proof
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By Non-Greatest Element of Well-Ordered Class has Immediate Successor, $\map \Succ i$ is guaranteed to exist.
Let $i \prec j$.
Let $S_i := \set {q \in S: i \prec q}$.
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Then $\map \Succ i$ is the minimal element of $S_i$.
By supposition, $j \in S_i$.
Thus:
- $j \nprec \map \Succ i$
Since Well-Ordering is Total Ordering:
- $\map \Succ i \preceq j$
Thus by supposition:
- $\map f {\map \Succ i} \le \map f j$
Since $\map f i < \map f {\map \Succ i}$:
- $\map f i < \map f j$
by transitivity.
$\blacksquare$