Example of a not regular topological space:Prove

Question

I am trying to prove that the following topology in $\mathbb{R}$ is not regular. It is defined as follows:

For $x\neq0$, a neighborhood base for $x$ is determined by $(x-\alpha,x+\alpha)$ with $\alpha>0$.

For $x=0$, a neighborhood base is $B_\alpha=\{t\in \mathbb{R}|-\alpha<t<\alpha,t\neq1,\frac{1}{2},\frac{1}{3}...\}$.

I have thought of taking the closed set $C=\{1,\frac{1}{2},\frac{1}{3}...\}$ as a counterxample of the definition of the regular space but I don't see how to showitformally. Appreciate some help. Thanks.

Answer 1

Yes, that's a good choice. Take $C$ and $0$. Let $V$ be a neighbourhood of $0$ and $U$ be a neighbourhood of $C$.

For some $r>0$ we have $(-r, r)\setminus C\subseteq V$. Now if you take $n$ large enough so that $1/n < r$, then we can find $s > 0$ with $(\frac1n-s, \frac1n+s)\subseteq U$., and clearly $((-r, r)\setminus C)\cap (\frac1n - s, \frac1n + s) \neq\emptyset$ since it's enough to take any $x$ with $\max(\frac1n-s, \frac1{n+1})< x < \frac1n$. So $U\cap V\neq\emptyset$. Since those were arbitrary neighbourhoods, this space isn't regular.

Answer 2

to show regularity, u need a closed subset and a point not in the closed set,what is the point that u are going to use?I think u mean to use zero.So let V be any nbd of zero, it contains a basic nbd of the form $B_{\alpha}$,now for $n\geq \frac{1}{[\alpha]+1}$,$\frac{1}{n} \in \left ]-\alpha,\alpha\right [$.if $V$ is any open subset containing $C$,then it contains necessarly a basic (euclidean) nbd $W$ of $\frac{1}{n}$,which can be assumed to be contained in $\left ]-\alpha,\alpha\right [$, hence $B_{\alpha} \cap V \neq \emptyset$(More precisely it contains the intersection of $W$ with the complement of $C$) .So our space isn'regular (${0}$ canot be seperated from $C$)