Lower limit topology on $\mathbb R$ is regular

Question

I want to prove that lower limit topology $\Bbb R_l$ is regular and am taking the approach as follows:

Let $A$ be a closed set and $x$ a point in $\Bbb R_l$ such that $x \notin A$.
Let $a \in A$. If $a >x$, then take the open set $[a, 2a)$ containing $a$ and if $a<x$, then take the open set $[a,q)$ containing $a$ where $q$ is any rational number between $a$ and $x$.

Now if we take the union of all these open sets for all $a \in A$ and denote it by $U_A$ then clearly, $U_A \cap \{x\}= \phi$ because each individual open set is disjoint with $\{x\}$.

But now I am stuck with finding an open set containing $x$ which does not intersect with $A$.

Answer 1

$A$ is closed so you already have a basic open set containing $x$ now that open set will be of the form $[x,y)$ and $[x,y)\cap A=\phi$.

Now choose that open sets $[a,q)$ in such a way that $[a,q)\cap [x,y)=\phi$. This is always possible as for every $a\in A$ either $a>y$ or $a<x$. So you are done.

Answer 2

Noob mathematician has shown how you can modify your approach slightly to get the desired result. That modification can actually be carried a lot further and allow you to avoid picking nbhds of individual points of $A$. As Noob mathematician said, since $x\notin A$, and $A$ is closed, there is a basic open nbhd $[x,y)$ of $x$ such that $[x,y)\cap A=\varnothing$. But the set $[x,y)$ is not only open in $\Bbb R_\ell$: it is also closed, so its complement $U=\Bbb R_\ell\setminus[x,y)$ is open in $\Bbb R_\ell$. And clearly $U\supseteq A$, so $[x,y)$ and $U$ are disjoint open sets containing $x$ and $A$, respectively.