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!
$\Bbb Q$

Added: $\{q\in\Bbb Q:q&lt;\sqrt2\}$ is a chain, it’s bounded above by $2$, say, and it has no least upper bound in $\Bbb Q$, because $\sqrt2$ is irrational.
One more answer, not discussed till now.

Consider $A = \{x,y,z\}$. So $\mathbb{P}(A)=  \{\phi, \{x\}, \{y\}, \{z\}, \{x, y\}, \{y,z\}, \{x, z\}, A\}$. 

Consider partial order relation $&lt;$ on $\mathbb{P}(A)$ as $a&lt;b$ iff $a\subset b$.

Consider the element $\phi \in \mathbb{P}(A)$. See nay element of $\mathbb{P}(A)$ is an upper bound of the set $\phi$. What is the least upper bound ? Not exists.