Suppose $X$ is a metric space. Let $C$ denote the collection of all dense subsets of $X$. Show that $\bigcap C = iso(X)$ Suppose $X$ is a metric space. Let $C$ denote the collection of all dense subsets of $X$. Show that $\bigcap C = \mathrm{iso}(X)$, where $\mathrm{iso}(X)$ refers to the set of all isolated points of $X$.
Attempt:
$C$ denotes the collection of all dense subsets of $X \implies \overline D = X ~~\forall~~D \in C ~~~\dots\dots (1)$.
Hence, $\mathrm{dist}(~x,D~)= 0~\forall~x \in X ~~~\dots\dots (2)$
Let $i \in X$ be an isolated point. Then : $\mathrm{dist} (~ i, X \backslash \{i\} ~)>0. ~~~\dots\dots (3).$
$D \backslash \{i\} \subseteq X  \backslash \{i\}  \implies \mathrm{dist} (~ i, D \backslash \{i\} ~) >0.~~~\dots\dots (4)$
Now, since $D$ is dense in $X$ and  $i \in X \implies \mathrm{dist} (~ i, D ~)=0$
$\implies i \in \overline D$
$\implies i \in D~ \bigcup~ acc ~D.~~~\dots\dots (5). ~\mathrm{acc} $ refers to the accumulation point
From $(4),(5):~i \notin \mathrm{acc}~D. \implies i \in D~~\forall~~D \in C$.
$\implies \mathrm{iso}~ (X) \subseteq \bigcap C~~~\dots\dots (6)$
Now,to prove the other way around :
Suppose $z \in \bigcap C.$ We need to show that $\mathrm{dist} (~ z, X \backslash\{z\} ~)>0$.
Suppose $z \notin \mathrm{iso}~X. \implies \mathrm{dist} (~ z, X \backslash\{z\} ~)=0$.
$\implies z\in \mathrm{Cl}~(X \backslash \{z\}).~\mathrm{Cl}$ refers to the closure.
Could someone please tell me how to move ahead?
Thank you very much for the help!
 A: That $\bigcap C\subseteq\operatorname{iso}(X)$ easily follows from the following:
Claim: If $x\in X$ is not isolated, then $X\setminus\{x\}$ is dense in $X$.
Proof: Suppose that $x\in X$ is not isolated. Let $V\subseteq X$ be a neighborhood of $X$. By definition, $V$ contains a point other than $x$, so $V\cap (X\setminus\{x\})\neq\varnothing$. This means that $x$ is an accumulation point of $X\setminus \{x\}$, so $x\in\operatorname{cl}(X\setminus\{x\})$. Therefore, $$X=\{x\}\cup(X\setminus\{x\})\subseteq\operatorname{cl}(X\setminus\{x\}),$$ which readily entails that $X\setminus\{x\}$ is dense. $\blacksquare$
A: By definition an element $x$ is isolated if $\left\{ x\right\} $
is an open set. 
Let $x$ be isolated. If $D$ is dense then $x\in\overline{D}$ so any open
set containing $x$ has a nonempy intersection with $D$. That results
in $\left\{ x\right\} \cap D\neq\varnothing$ or equivalently $x\in D$.
Conversely if $x$ is an element of each dense set then the set $E:=X-\left\{ x\right\} $
is not dense. Then $\overline{E}\neq X$ or equivalently $x\notin\overline{E}$.
So an open set $U$ exists with $x\in U$ and $U\cap\left(X-\left\{ x\right\} \right)=\varnothing$.
That tells us that $U=\left\{ x\right\} $, hence that $\left\{ x\right\} $
is an open set. Proved is now that $x$ is isolated.
