Let $S \subseteq \mathbb{R^n}$ and denote the set of limit points of
$S$ as $S'$.Show that $S \cup S' = \bar{S}$ where $\bar{S}$ is the closure.
(i) I want to show that $\bar{S} \subseteq S \cup S'$ .
Let $x \in \bar{S}$ then either $x \in S$ or $x \in \bar{S}\backslash S $.
If $x \in S$ then $x \in S \cup S'$
If $x \in \bar{S}\backslash S $ …?
(ii) I want to show that $S \cup S' \subseteq \bar{S}$.
Let $x \in S \cup S'$. Then, $x \in S$ or $x \in S'$.
If $x\in S$ then $x \in \bar{S}$.
If $x \in \bar{S}$ … ?
Thank you
Best Answer
Looking at (i) you're almost done. If $x \in \overline{S} \backslash S$ then $x \notin S$ and since $x \in \overline{S}$ then all neighborhoods $U$ of $x$ intersect $S$. But since $x$ is not in $S$ then $U$ must intersect $S$ at some point other point that is not $x$. Thus $x \in S'$. Therefore you can conclude that $\overline{S} \underline{\subset} S \cup S'$.
For (ii) note that if $x \in S \cup S'$ then all neighborhoods $U$ of $x$ intersect $S$ at some point, doesn't matter where. Thus $x \in \overline{S}$. Therefore $S \cup S' \underline{\subset} \overline{S}$.
The final conclusion being that $S \cup S' = \overline{S}$
(This is a general result in topology that you can reformulate to real euclidean space if you wish)