Is there example of two disjoint closed sets in Rational Sequence topology. The set of the real numbers with the Rational Sequence Topology is not normal by Jone's Lemma. What is a sample of two disjoint closed sets that cannot be separated by two disjoint open sets? 
 A: Unfortunately, explicit examples cannot be given.
First I’ll describe the space in order to establish some notation. Let $\Bbb P$ be the set of irrational numbers. For each $x\in\Bbb P$ let $\langle q_k^x:k\in\Bbb N\rangle$ be a sequence of rationals converging to $x$ in the usual topology on $\Bbb R$. For each $n\in\Bbb N$ let $B(x,n)=\{x\}\cup\{q_k^x:k\ge n\}$. Let $X$ be the set of real numbers with the following topology: points of $\Bbb Q$ are isolated, and for each $x\in\Bbb P$, $\{B(x,n):n\in\Bbb N\}$ is a local base of open sets at $x$. The space $X$ with this topology is the rational sequence space. Clearly every subset of $\Bbb P$ is closed in $X$.
We know from Jones’s Lemma that there are disjoint subsets $H$ and $K$ of $\Bbb P$ that cannot be separated by disjoint open sets. However, we cannot actually identity specific $H$ and $K$ with that property unless we know which sequences $\langle q_k^x:k\in\Bbb N\rangle$ were used to construct the topology. The reason is that if $H$ and $K$ are any two disjoint subsets of $\Bbb P$, we can choose the rational sequences in such a way that $H$ and $K$ can be separated by disjoint open sets in $X$. Specifically, partition $\Bbb Q$ into two dense sets $Q_H$ and $Q_K$, and choose the rational sequences so that $q_k^x\in Q_H$ for each $x\in H$ and $k\in\Bbb N$, and $q_k^x\in Q_K$ for each $x\in K$ and $k\in\Bbb N$. Then $\bigcup_{x\in H}B(x,0)$ and $\bigcup_{x\in K}B(x,0)$ are disjoint open nbhds of $H$ and $K$, respectively.
A: Here's an example for the special choice $q_n^x=[nx]/n$: let $A$ and $B$ be
disjoint dense (in the standard topology) subsets of the irrationals which are complementary and $U\supset A$ $V\supset B$ be open. Futhermore, let
$A_n=\{x\in A:q_k^x\in U, k\ge n\}$ and $B_n=\{x\in B: q_k^x\in V, k\ge n\}$.
By Baire's category theorem, there is an $n$ such that $A_n$ or $B_n$ is
dense (again in the standard topology) in some open interval $I$, assume wlog
that it is $A_n$. Then $U$ contains all the rationals in $I$. But $B$ is also
dense in $I$, so $V$ must also contain some rational in $I$, and so $U$ and $V$ cannot be disjoint.
