The neighborhood of each points of a special topological space Let $R$ is the real line and $R^* = R \cup \{x^*\}$, where $x^*$ is not in $R$. For any subset $A$ of $R^*$, we define the $cl_{R^*}(A)$as following:
if $A$ is finite, then $cl_{R^*}(A)=A$; if $A$ is infinite, then $cl_{R^*}(A)=cl_R(A-\{x^*\}) \cup \{x^*\}$.
This generates a topology on $R^*$.
My question is this: What's the nbhd of $x^*$ and the nbhd of $x$ of $R$?
 A: If $A$ is an infinite subset of $\Bbb R$, then $x^*\in\operatorname{cl}_{\Bbb R^*}A$, so $\Bbb R^*\setminus A$ does not contain an open nbhd of $x^*$. On the other hand, if $F$ is a finite subset of $\Bbb R$, then $\Bbb R^*\setminus F$ is an open nbhd of $x^*$. It follows that the open nbhds of $x^*$ are precisely the sets $\Bbb R^*\setminus F$ such that $F$ is a finite subset of $\Bbb R$: the topology at $x^*$ is cofinite.
Now suppose that $x\in\Bbb R$, and let $U$ be an open nbhd of $x$ in $\Bbb R^*$. If $x^*\in U$, then $U=\Bbb R^*\setminus F$ for some finite $F\subseteq\Bbb R\setminus\{x\}$, so assume that $x^*\notin U$. Then $\Bbb R^*\setminus U$ is a closed set containing $x^*$. Let $C=\Bbb R^*\setminus U$. Then
$$C=\operatorname{cl}_{\Bbb R^*}C=\operatorname{cl}_{\Bbb R}\big(C\setminus\{x^*\}\big)\cup\{x^*\}\;,$$
so $C\setminus\{x^*\}=\operatorname{cl}_{\Bbb R}\big(C\setminus\{x^*\}\big)$, and $C\setminus\{x^*\}$ is therefore closed in $\Bbb R$. But $C\setminus\{x^*\}=\Bbb R\setminus U$, so $U$ is open in $\Bbb R$. Thus, $U\subseteq\Bbb R^*$ is an open nbhd of $x\in\Bbb R$ iff either $U\subseteq\Bbb R$ is an open nbhd of $x$ in $\Bbb R$, or $U=\Bbb R^*\setminus F$ for some finite $F\subseteq\Bbb R\setminus\{x\}$.
(I normally try not to give a complete answer right away when the question is homework, but in this case (a) I didn’t notice the tag until after I’d posted the answer, and (b) I’m not at all sure that I could give a very helpful hint without doing most of the problem anyway.)
