Not well-ordered sets satisfying the least upper bound property Before proving that every closed interval in $\mathbb{R}$ is compact, James R. Munkres remarks that, in Section 27 entitled "Compact Subspaces of the Real Line" of Topology (2nd edition),

Remark: We need only one of the order properties of the real line --- the least upper bound property. We shall prove the theorem using only this hypothesis; then it will apply not only to real line, but to well-ordered sets and other ordered sets as well.

The theorem mentioned above is as follows.

Theorem 27.1 Let $X$ be a simply ordered set having the least upper bound property. In the order topology, each closed interval in $X$ is compact.

According to the remark, the least upper bound property seems to be a subtle property of order sets, and 

(Problem:) there should be some ordered sets which are not well-ordered while satisfying the least upper bound property. Could you give some examples as such?

Any comments on the relation between ordered sets and the least upper property are appreciated. 
 A: A nice example is $\mathbb{N} - \{ 0 \}$ ordered by divisibility, i.e. $a \le b$ if and only if $a \mid b$. This is a partial order which is not linear (let alone well-ordered).
If a set $X \subseteq \mathbb{N} - \{ 0 \}$ has an upper bound in this order then there must be some $a$ into which all $x \in X$ divide; in particular $X$ must be finite, and its least upper bound is the least common multiple of the elements of $X$.
A: The least upper bound property just means that given a bounded set we can specify a unique bound, which is also the smallest possible bound.
Well ordered sets have this property because every non-empty set has a least element, so given a bounded set, the set of its upper bounds is non-empty and has a least element.
But the same holds for "anti-well ordered sets", that is sets whose reverse order type is well-ordered, e.g. the negative integers. Combining this, you can even show that the integers themselves which are neither of the two, have the least upper bound property.
But not every ordered set has the least upper bound property, for example consider $\Bbb R\setminus\{0\}$, then the set of negative real numbers is bounded, but has no least upper bound. Similarly $\Bbb Q$ itself is missing "a lot of least upper bounds".
However, every linear order can be extended to a linear order which satisfies the least upper bound property by considering a construction similar to the construction of the real numbers by a Dedekind completion of the rationals.
