# Talk:Limit Points of Countable Complement Space

In this page, it is proven that if $H\subseteq S$ is countable or finite, then it contains all its limit points, but the result is much stronger: $H$ has no limit points.
Fix $x\in S$. Consider $U=\left(S\setminus H\right)\cup\left\{x\right\}$. $U$ is open in $S$, as $S\setminus\left(\left(S\setminus H\right)\cup\left\{x\right\}\right)=H\setminus\left\{x\right\}$ is countable or finite. Hence $U$ is a neighborhood of $x$. The definition of limit point states then that $x$ is not a limit point.
The result in the page isn't wrong as if $H$ has no limit points, they all (none) lay in $H$.