Proving intersection of dense subsets of a metric space X is the isolated points of X. Suppose X is a metric space. Let $\mathscr C$ denote the collection of all dense subsets of X. Show that $\bigcap\mathscr C $ = iso(X).
Thus the question asks to prove that every dense subset of X only contains all the isolated points of X. I want to prove this by contradiction but do the proof.....please help.
I found the problem in one exercise of Searcoid's Book on Metric Spaces.
Thank You!!
 A: If $x$ is an isolated point of the space $X$ then any dense subset of $X$ contains $x$, so 
$ iso(X)\subset\bigcap\mathscr C$. From the other side, if $x$ is not an isolated point of the space $X$, then the set $X\setminus \{x\}$ is dense in $X$, so $x\not\in\bigcap\mathscr C$.
A: Let $x$ be an isolated point and $D$ a dense subset. Then $V=\{x\}$ is open.
Now $D\cap V\neq \emptyset$ , hence $x\in D$.
A: 
the question asks to prove that every dense subset of $X$ only contains all the isolated points of $X$

Not quite. The question asks to prove that the only points common to all the dense subsets of $X$ are the isolated points. The set $\text{isol}(X)$ of isolated points of $X$ will not be dense unless $X$ is discrete. For example, 
the subsets 
$$\{0\}\cup\left(\mathbb{Q}\cap[1,2]\right)$$
and 
$$\{0\}\cup\left((\mathbb{R}\setminus\mathbb{Q})\cap[1,2]\right)$$
of the metric space $\{0\}\cup[1,2]$ are both dense. 
Both contain many more points than just $0$ (which is the sole isolated point in this space/in this case).
To answer your question, we will prove that $\bigcap\mathscr{C}\supseteq\text{isol}(X)$ and that $\bigcap\mathscr{C}\subseteq\text{isol}(X)$, from which it will follow that $\bigcap\mathscr{C} = \text{isol}(X)$.
First recall that $x$ is isolated iff $\{x\}$ is open iff $X\setminus\{x\}$ is closed.
If two sets are disjoint, then one is contained in the complement of the other. 
Thus if $x$ is isolated and $D\cap\{x\}$ is empty, then $D\subseteq X\setminus\{x\}$, 
which means $D$ cannot be dense, 
because its closure cannot exceed the closed set $X\setminus\{x\}$.
So, every dense set intersects $\{x\}$, which means every dense set contains $x$; since $x$ was an arbitrary isolated point, we conclude that every dense set contains every isolated point: $$\bigcap\mathscr{C}\supseteq\text{isol}(X).$$
Now, if $x$ is not isolated, then $X\setminus\{x\}$ is not closed, so its closure is strictly bigger: its closure is the whole space $X$. Therefore, each non-isolated point $x$ is avoided by some dense set, namely $X\setminus\{x\}$.
Thus, the intersection of all the dense sets avoids all the non-isolated points: $$\left(X\setminus\text{isol}(X)\right) \cap \bigcap\mathscr{C} = \varnothing.$$
But, again, if two sets are disjoint, then one is contained in the complement of the other, so $$\bigcap\mathscr{C} \subseteq \text{isol}(X)$$
and we're done.
Note. $X$ need not be a metric space; the above proof holds in any topological setting.
