Give an example of a simply ordered set without the least upper bound property. In Theorem 27.1 in Topology by Munkres, he states "Let $X$ be a simply ordered set having the least upper bound property.  In the order topology, each closed interval in $X$ is compact."
(The LUB property is if a subset is bounded above, then it has a LUB.)
I don't understand how you could have a simply ordered set (a chain) WITHOUT the LUB property.  If a subset is bounded and it is a chain, then how can it not have a LUB?
Can someone give an example?
Thanks!

According to the strict definitions given by the OP, the null set fails to have a Least Upper Bound while still being simply ordered.
The Least Upper Bound of a set, as defined at the Wikipedia page he links to requires that it be a member of that set.  The null set, having no members, clearly lacks a LUB.
However, the definition given for being simply ordered does not require that the set have any elements.  Indeed, a set can only lack the property if it has a pair of elements that are not comparable.
So, the null set is indeed Simply Ordered without having the Least Upper Bound property. 
