Munkres Topology Question article 17 problem 5 Let $X$ be an ordered set with the order topology. Show that $\overline {(a,b)}\subset [a,b]$.Under what conditions does equality hold?
Proof:Since $\overline{(a,b)}$ is the smallest closed set containing $(a,b)$ and $[a,b]$ is a closed containing it so the relation follows
But how to show the equality part?

Here's my idea (feel free to criticize): 
if $a = b$, then $(a,b) = \emptyset$ which means $\overline{(a,b)} = \emptyset \neq [a,b] = \{a \}$.  
But if $a \neq b$, you could still have $(a,b) = \emptyset$ if there is no $z \in X$ with $a < z < b$.  
So, what we really want is for there to exist $z \in X$ such that $a < z < b$ (note that this implies $a \neq b$).  But then, if $a$ is not the smallest element of $X$, and there is only one $z$ such that $a < z < b$, then we would have an open neighborhood around $a$ (namely, $(a', z)$ where $a' < a$) such that the neighborhood doesn't intersect $(a,b)$, which would mean $a$ is not a limit point of $(a,b)$...
I think what we need is the ability to always find a point between two points in the space, that is, if $x, y \in X$ we need to be able to find $z \in X$ such that $x < z < y$.
