Suppose $(A,\preccurlyeq)$ is a well-ordered chain, i.e., every subset $B\subseteq A$ has a least element under the order $\preccurlyeq$.
We say that a set $B\subseteq A$ is bounded if there exists $s\in A$ such that $b\preccurlyeq s$ for all $b\in B$.
Question: Does every bounded subset of $A$ have a maximum?
If $B$ is finite this is obviously true, since we can start with the minimum $m=\min B$, and keep applying the successor operation
$$x^+=\min\textstyle_{\preccurlyeq}(B\smallsetminus\{b\in B: b\preccurlyeq x\})$$
and stop when we end up with a singleton $\{M\}$, this will be the maximum of $B$. But what if $B$ is infinite?
I can't think of an easy counterexample to this, it feels like it should be true.
Best Answer
If we let $A = \{-\frac{1}{n} \mid n \in \mathbb{N}_+\} \cup \{0\}$ with the usual order, then $A$ is well-ordered. Now take $B = A \setminus\{0\}$ which is bounded, as $0$ is an upper bound for $B$, but as $0$ is not an element of $B$, and no other element of $A$ provides an upper bound for $B$, it follows that $B$ has no maximum.