Let $(X, d)$ be a metric space and $G ⊆ X$. and prove the following Let $(X, d)$ be a metric space and $G ⊆ X$. Show the following:
(i) If $x ∈ G$ then $x$ is either a limit or isolated point of $G$.
(ii) If $G$ is finite then $G$ is closed.
(iii) The set of limit points of $G$ is closed.
I'm not quite certain how to approach these problems even though I've tried using the definitions of the different points. Thank you for your help!
 A: For (i), if $x$ is not a limit point of $G$ that means there is some open $x\in U$ such that $U\cap G =\lbrace x\rbrace$.For (ii), a point in a metric space is a closed set.  A finite set is an union of (its) points and since closed sets are closed under finite unions, a finite set is closed. For (iii),  suppose that $x_n\rightarrow x$ and each $x_n$ is a limit point. Take any open set $x\in U$ so that there is some $p$ such that for all $m\geq p$, $x_m\in U$. Now $U$ is also an open set containing limit points (elements from $x_n$) and thus it follows that $x$ is a limit point itself and hence, the set of limit points is closed. Note we could assume $x_n\neq x$ without loss of generality.
Maybe (iii) a bit more clearly : Let $L$ be the set of limit points of $G$. We want to prove that $L=\overline{L}$. So take $x\in\overline{L}$ so that for every open $x\in U$ one has that $U\cap L\neq \emptyset$. Let $y\in U\cap L$ and take open $V$ such that $y\in V\subseteq U$ and such that $x\notin U$, which is possible since $X$ is Hausdorff. Now, $V\cap G\setminus \lbrace y\rbrace \neq \emptyset$ and thus, $x\in L$.
