Give an example of a simply ordered set without the least upper bound property.

Solution 1:

$\Bbb Q$

Added: $\{q\in\Bbb Q:q<\sqrt2\}$ is a chain, it’s bounded above by $2$, say, and it has no least upper bound in $\Bbb Q$, because $\sqrt2$ is irrational.

Solution 2:

Consider $(-1,0)\cup(0,1)$ with the natural ordering (which is linear). $(-1,0)$ has no least upper bound although it is bounded (eg. by $1/2$).

Solution 3:

$\{0,1\}\times\Bbb Z$ ordered lexicographically. The subset $\{0\}\times\Bbb N$ is bounded from above but has no least upper bound.

Solution 4:

Up to isomorphism, the minimal example is the set $$ \left\{-\frac{1}{n}\;;\;n\geq 1\right\}=-\frac{1}{\mathbb{N}^*}\quad\mbox{in}\quad\frac{1}{\mathbb{Z}^*}=\left\{\frac{1}{n}\;;\;|n|\geq 1\right\} $$ with the obvious ordering.

Solution 5:

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.