Is the following statement true: $S$ is closed in a metric space if every point in $S$ has some sequence converging to it Suppose that $M$ is a metric space containing $S$,
then is it true that:

$S$ is closed (complement is an open set) if every point in $S$ has
  some sequence converging to it

I am not certain because I have only seen the reverse statement, i.e., let $ \{a_n\}_{n = 1}^\infty$ be a sequence of elements in $S$, then $S$ is closed if the limit of any such sequence is in $S$. 
If the statement is true, wouldn't all we have to do is to take the constant sequence $p,p,p,p,p,p,p, \ldots$ to show that $S$ is a closed set?
 A: No, this is not true. 
Consider $S=(0,1)$ as a subset of $\Bbb R$ in the usual topology. $S$ is not closed (as $0$ and $1$ are adherent points not in $S$) but every point in $S$ has a seuence (from $S$ even) converging to it. In fact if we have any subset $S$ of the reals, and any $x \in S$, $x+\frac{1}{n} \to x$, so the property of there being a sequence in $\Bbb R$ converging to $x$ has noting to do with $S$, but holds for all points, simply because there are no isolated points in the space. Or it holds in any space if we allow constant sequences (just define $x_n =x$ always)! 
But even the mild property "for all $x \in S$ there is a sequence (with infinitely many terms) in $S$ that converges to $x$" only says that no point of $S$ is an isolated point, and this can hold for many non-closed sets too. 
The correct fact is (for metric spaces certainly) : $S$ is closed iff "for every sequence $(x_n)$ from $S$ such that $x_n$ converges to $x$, we know that $x \in S$", so you cannot "walk out of $S$ with a sequence from $S$", informally. 
This we can see in my first example, which fails this property, as e.g. $\frac1n \in S$ but its limit $0 \notin S$. 
