Does there exist an well ordered set $(A,<)$ that isn't topologically complete, i.e. that for every $$B\subseteq A,~ B\not = \emptyset$$ such that $B$ is bounded above with some $c\in A$, $B$ has supremum in $A$.
$\mathbb N$ is obviously complete since it is locally finite. And ideas such as $\mathbb N\times \{0,1\}$ or $\left(\mathbb N\times\{0\}\cup \mathbb Z\times\{1\}\right)$ dont work either.
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Actually i think i proved it. Given $B \subseteq A$ defined as above, we have some $c \in A$ such that $c \ge b, \forall b \in B$. Now we know that set of all upper limits of $B$ is non-empty, therefore, it contains the minimum $m$. It is easily seen that $m$ is the supremum of $B$ since it is the smallest upper limit of $B$.
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Rephrasing your definition of topological completeness to note that it's exactly the same as the least upper bound property, we note that any well-order has that property.
This has the standard proof: let $C$ be the set of upper bounds of $B$. It's non-empty by definition, and so by the well-order property has a least element.
This least element is the supremum of $B$.
No, there is no such well-ordered set. Suppose $(A, \le)$ is well-ordered, and $\emptyset \ne B \subseteq A$ with $B$ bounded above by some element of $A$. Then the set of upper bounds of $B$ is a nonempty subset of $A$, and so has a least element $b_0$. It's easy to see that $b_0 = sup(A)$.