"Closed interval" on ordered topology 
Let $X$ be linearly ordered by a relation $\leq$. Taking as a subbase for topology on $X$ all sets of the form $\{x;x<a\}$ and $\{x;x>a\}$, for $a\in X$. Can be $\{x\in X; a\leq x\leq b\}$ a open set, for $a<b$, in $X$?

I think the answer is no, beause $\{x\in X; a\leq x\leq b\}^c=\{x;x<b\}\cup\{x;x>a\}$ is a open set.
Its correct?
 A: The answer to your question is "Yes, $\{x\in X;\;a\leq x\leq b\}$ for $a<b$ can be an open set."
There are a number of ways this can happen. The trivial but slightly boring one is if your order has endpoints $a$, and $b$ such that $\forall x\in X\; a\leq x$ and $\forall x\in X\; b\geq x$. Then $X=\{x;a\leq x\leq b\}$ and is open. An example of this is $[0,1]$ with the order inherited form $\mathbb{R}$.
A less trivial example is any linear order which is at least partially discrete. This just means you have a few points such that there is nothing between them. A small example would be $\{1,2,3,4,5\}$ ordered the usual way (with order inherited from $\mathbb{N}$. In this case every singleton is open. We will just quickly show it for $\{2\}$. We can take the open sets $\{x;x>1\}$ and $\{x;x<3\}$ their intersection is open and is exactly $\{2\}$. Since the union of open sets is open every set in this topology is open (it is the discrete topology) and thus in particular the set $\{x;1\leq x\leq 2\}$ is open for example.
This second argument can be extended to arbitrary linear orders for which the ray topology gives us a discrete topological space (I believe these are exactly the discrete linear orders but I might be wrong something strange might happen with tricky enough orders like say $\omega\times \mathbb{Z}$). Certainly an easy example of a bigger space which is discrete under the ray topology is $\mathbb{Z}$ and for it all the sets you describe will be open.
