Context: trying to get someone else to do my thinking for me.
Reference: Smullyan and Fitting: Set Theory and the Continuum Problem (rev ed. 2010)
Chapter $4$: Superinduction, Well-Ordering and Choice: $\S 1$: Introduction to Well-Ordering
We are looking at Exercise $1.2$.
For a linear ordering $\le$ on a class $A$, a subclass $L$ of $A$ is called a lower section of $A$ (with respect to $\le$, understood) if every element of $L$ is less than every element of $A$ not in $L$. A lower section $L$ of $A$ is called a proper lower section if it is non-empty and not the whole of $A$.
Prove that a sufficient condition for a linear ordering $\le$ 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 $L$.
So, suppose $A$ is a class under $\le$ such that $\le$ has exactly the above property.
To demonstrate a well-ordering it is sufficient to show that every non-empty subclass of $A$ has a least element. Clearly $A \setminus L$ has a least element. But that's just $A \setminus L$. I can't get further by abstract thinking on the matter.
So I thought: what if I try to create a counterexample? What about the set of integers $\Bbb Z$ under the usual ordering? That's a linearly ordered set (and a set is a class innit).
Let $x$ be some element in $\Bbb Z$. Then the set $\{y \in \Bbb Z: y < x \}$ is a lower section of $\Bbb Z$ which has the above properties. That is, the element $x$ is itself that smallest element of $A \setminus L$.
Are there any other proper lower sections of $\Bbb Z$ which are not of the form $\{y \in \Bbb Z: y < x \}$? I can't see how such a proper lower section could be constructed.
But while it appears as though $\Bbb Z$ fulfils the conditions for $(\Bbb Z, \le)$ to be a well-ordered set, $\Bbb Z$ most patently is not a well-ordered set, because it does not have a smallest element.
There must be something I am missing.