Condition on Proper Lower Sections for Total Ordering to be Well-Ordering/Mistake

Source Work

 * Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering:
 * $\S 1$ Introduction to well ordering:
 * Exercise $1.2$
 * Exercise $1.2$

Mistake

 * Prove that a sufficient condition for a linear ordering $\le$ of $A$ to be a well ordering of $A$ is that, for every proper lower section $L$ of $A$, there is a least element $x$ of $A$ not in $L$.

Analysis
Consider the integers $\Z$ under the usual ordering $\le$.

We have that $\struct {\Z, \le}$ is a linearly ordered set.

Yet let $x \in \Z$ be arbitrary.

We have that $\set {y \in A: y < x}$ is a proper lower section of $\struct {\Z, \le}$.

There is no other way to construct a proper lower section of $\struct {\Z, \le}$.

Then we note that $x$ is then the least element of $A$ not in $L$.

Hence $\struct {\Z, \le}$ fulfils the conditions of the hypothesis, but is not a well ordered set, as can be seen by the fact that there is no smallest integer.

It is suggested that necessary condition was meant, not sufficient condition.

If so, the proof is trivial.