Prove that the space $\Bbb R_K$ is not regular.

Question

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 $\Bbb R_K$ .

2. $0\notin A$.Claim $0,A$ can't be strongly separated.
3. 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

Answer 1

Hint: There is some $0\in (-x,x)-K \subset U$. Choose $N\in \Bbb{N}$ such that $\frac{1}{N} \in (-x,x)$. Now $\frac{1}{N}$ is in $V$. Can you see why $V$ must intersect $(-x,x)-K$ (and thus $U$) non trivially from that?

Answer 2

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 $[U\cap V]\supset (d,c)\backslash K\ne \phi.$