Say a subset $S \subset \mathbb{R}$ has $\textbf{least-upper-bound property (LUBP)}$ if every nonempty bounded above subset $T \subset S$ has supremum that exists in $S$.
I want to characterize the LUBP subsets of $\mathbb{R}$. I know that all closed intervals and singeltons have LUBP, and moreover, the union of two LUBP sets is LUBP, since if $S = A \cup B$ and $T \subset S$, then $\sup T = \max(\sup(T \cap A),~\sup(T \cap B))$), so certainly any finite union of closed intervals and singeltons have LUBP, but then we're still missing sets like $\mathbb{Z}$.
Any ideas?
(After the comment of Paul Sinclair). Consider the topology on $\mathbb R$ generated by intervals of the form $(a,b]$.
Claim: The closed sets are the subsets of $\mathbb R$ with the least upper bound property.
Proof: Consider a closed set $C$ and a subset $T\subset C$ so that $T$ has an upper bound $M$, then $t\le M$ for all $t \in T$. Let $m = $ the supremum of all elements of $T$, we know that $m$ exists as a real number as long as $T$ is nonempty. Consider $m\in (a,b]$, then we have $a<m$ and by definition of $m$ there must exist $t \in T$ such that $a<t\le m$, so $(a,b]\cap C \neq \emptyset$. It follows that $m\in C$.
Finally, let $C$ be a subset of $\mathbb R$ with the least upper bound property and consider $x\in \mathbb R\setminus C = U$. Let $T=\{c\in C \vert c\le x\}$. If $T$ is empty then $(a,x]\subset U$ for any $a<x$. If $T$ is non empty, observe that $x$ is an upper bound for $T$ and therefore the supremum of $T$, call it $s$, is also in $C$. In this case, we must have $s<x$, and therefore $(s,x]\subset U$. It follows that $C$ is closed.