Prove that the space $\Bbb R_K$ is not regular.
where the basic open sets on $\Bbb R_K$ is given by $\{(a,b):a,b\in \Bbb R\}\cup \{(a,b)-K\}$ where $K=\{\dfrac{1}{n}:n\in \Bbb Z_+\}$.
[Hint: Look at $0$ and $\{\dfrac{1}{n}:n\in \Bbb Z_+\}$]
My try:
1.Let us consider the set $A=\{\dfrac{1}{n}:n\in \Bbb Z_+\}$ which is closed as it is the complement of $(0,1)-K$ and $(0,1)$ is open in $\Bbb R_K$ .
- $0\notin A$.Claim $0,A$ can't be strongly separated.
- Suppose they can be separated strongly then there exists disjoint sets $U,V$ such that $0\in U,A\in V$ such that $U \cap V=\emptyset$.Then obviously $U$ is not of the form $(a,b)$ otherwise it would contain members from $A$.Then $U$ has form say $(a,b)-K$
But $V$ can have a form of $(a,b)$
But how can I arrive at a contradiction from here? I think I have to somehow show that $U\cap V\neq \emptyset $.But I am stuck here.Any help
Best Answer
First, $U$ and $V$ are not necessarily members of the base, although we can take $U$ to be a member of the base. $V$ is a union of a family of members of the base.
Secondly, observe that for some $r>0$ we must have $U\supset (0,r)\backslash K,$ but there exists $n\in N$ with $1/n<r$. And since $1/n\in K,$ there must exist a member $C$ of the base with $1/n\in C\subset V.$
So $C=(a,b)$ with $a<b.$
Since $a<1/n<r,$ and $a<1/n<b,$ let $c= \min (r,b),$ noting that $c>a.$
Let $d=\max (0,a)$ noting that $d<c$. We have $[U\cap V]\supset (d,c)\backslash K\ne \phi.$