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.