TITLE: Show that if $U$ is a nonempty open set in $\mathbb{R}$, then $\overline{U}$ = $\mathbb{R}$.
QUESTION [5 upvotes]: Let $\mathbb{R}$ be a topological space with topology $T = \{ U \subseteq \mathbb{R} : U^c \ \text{is finite} \} \bigcup  \{\emptyset \}$. Show that if $U$ is a nonempty open set in $\mathbb{R}$, then $\overline{U} = \mathbb{R}$. 

So I suppose that $U$ is a non open set in $\mathbb{R}$ 
$\Rightarrow$ $U$ is a subset of $\Bbb R$ such that U$^c$ is finite
What is the next step?

REPLY [2 votes]: If $U \in T\setminus\{\emptyset\}$, then 
$$
\mathbb{R}\setminus U=\{u_1,\ldots,u_m\}
$$ 
It follows that
$$
U=\mathbb{R}\setminus\{u_1,\ldots,u_m\}.
$$
To prove that $\overline{U}=\mathbb{R}$ it is enough to show that given $x \in \mathbb{R}$, we have $U_x\cap U\ne \emptyset$ for every $U_x \in T$ with $x \in U_x$. So let $U_x \in T$ with $x \in U_x$. Then $U_x=\mathbb{R}\setminus\{x_1,\ldots,x_n\}$. We now have
$$
U_x\cap U=(\mathbb{R}\setminus\{x_1,\ldots,x_m\})\cap(\mathbb{R}\setminus\{u_1,\ldots,u_m\})=\mathbb{R}\setminus(\{x_1,\ldots,x_n\}\cup\{u_1,\ldots,u_m\}).
$$
It is obvious that 
$$
\mathbb{R}\setminus(\{x_1,\ldots,x_n\}\cup\{u_1,\ldots,u_m\})=\mathbb{R}\setminus\{x_1,\ldots,x_n,u_1,\ldots,u_m\}
$$
is non-empty since $\{x_1,\ldots,x_n,u_1,\ldots,u_m\}$ is a finite subset of $\mathbb{R}$.