$[a,b]=X-((-\infty,a)\cup(b,+\infty))$ is closed and contains $(a,b)$ , so it contains the closure of $(a,b)$ . It equals the closure iff both endpoints are limit points of the interval, i.e. if $(a,b)$ is not empty and for every $x\in(a,b)$ there are $s,t\in(a,b)$ such that $a<s<x<t<b$ . This is equivalent to the requirement that $a$ has no immediate successor, and $b$ has no immediate predecessor. Otherwise, if $a$ has an immediate successor $c$ then $(-\infty,c)$ is an open set containing $a$ that does not intersect $(a,b)$ , and, similarly, if $b$ has an immediate predecessor $c$ then $(c,+\infty)$ is an open set containing $b$ that does not intersect $(a,b)$ .