Why is the topology on $\mathbb{R}$ formed by the basis $[a,b)$ normal? Why is the topology on $\mathbb{R}$ formed by the basis $[a,b)$ normal? I need to prove that two disjoint closed sets are contained wtihin two open disjoint sets. First, I tried to understand how a closed set looks in this topology, and so I took the complement of a base set, which is $(-\infty,a) \cup [b,\infty)$. Is it right to say that every closed set is the union of such sets? If this is the case it seems as though no two closed sets can be disjoint, therefore proving that the topology is normal in an empty way. I think I'm wrong somewhere here, but where?
Any help would be welcomed!
 A: No, it’s not true that every closed set in $\Bbb R$ with the lower limit topology is a union of sets of the form $\Bbb R\setminus[a,b)$. For example, $\{0,1\}$ is a closed set, because its complement is the open set
$$\bigcup_{x<0}[x,0)\cup\bigcup_{0<x<1}[x,1)\cup\bigcup_{x>1}[x,x+1)\;.$$
Let $\tau$ be the lower limit topology. It’s not hard to show that every $U\subseteq\Bbb R$ that is open in the usual topology belongs to $\tau$. Thus, if $C$ is any subset of $\Bbb R$ that is closed in the usual topology, $\Bbb R\setminus C$ is open in the usual topology and therefore in $\tau$, and it follows immediately that $C$, its complement, is closed in the lower limit topology. Thus, every set that’s closed in the usual topology is closed in the lower limit topology as well. There are also many sets that are closed in the lower limit topology but not in the usual topology, including all sets of the form $[a,b)$ with $a<b$: 
$$\Bbb R\setminus[a,b)=\bigcup_{x<a}[x,a)\cup\bigcup_{x>b}[b,x)$$
is open, so $[a,b)$ must be closed.
Here’s an extended hint for proving normality. Suppose that $H$ and $K$ are disjoint closed sets in $\langle\Bbb R,\tau\rangle$.


*

*Show that for each $h\in H$ there is an $x_h>h$ such that $[h,x_h)\cap K=\varnothing$.  

*Show that for each $k\in K$ there is an $x_k>k$ such that $[k,x_k)\cap H=\varnothing$.  

*Show that if $h\in H$ and $k\in K$, then $[h,x_h)\cap[k,x_k)=\varnothing$.  

*Let $U=\bigcup_{h\in H}[h,x_h)$ and $V=\bigcup_{k\in K}[k,x_k)$; show that $U$ and $V$ are disjoint open nbhds of $H$ and $K$, respectively.

A: Open sets are going to look like arbitrary unions of $[a,b)$. Closed sets are defined to be the complement of an open set. So if $U$ is open, $$U = \bigcup_{i \in I} {[a_{i}, b_{i})}.$$
Then,
$$ X - U = X - \bigcup_{i \in I} {[a_{i},b_{i})} = \bigcap_{i \in I} {[a_{i},b_{i})^{C}}$$
So your closed sets are looking like intersections of the sets you mentioned above. 
For example, the sets $(-\infty, a)$ and $[b, \infty)$ are both closed in this topology. 
