Well-Ordering is not necessarily Usual Ordering
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Theorem
According to Zermelo's Well-Ordering Theorem, $S$ can be well-ordered.
However, the usual ordering on $S$ may not necessarily be a well-ordering.
Proof
From Rational Numbers are Well-Orderable, it is possible to apply a well-ordering to the set of rational numbers $\Q$.
However, the usual ordering on $\Q$ is not a well-ordering.
Indeed:
- $\set {x \in \Q: x \le 0}$
has no smallest element.
$\blacksquare$
Sources
- 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 1$ Introduction to well ordering: Discussion