# Talk:Limit Points of Countable Complement Space

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## Result can be stronger

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 $H$.

- I've checked the source work and it appears to differ in presentation for what I put here. I'm going to need to address this at another time as I need headspace for this which I don't got at the moment.

- BTW please remember to sign your posts. --prime mover (talk) 17:02, 5 July 2013 (UTC)