If a set contains all its limit points must it be closed? If a set $X$ in a topological space $T$ has the property that for all sequences $x_n \in X, x_n \to x \implies x\in X$ must X be closed?
I know this is true for metric spaces but is it true for a general topological space.
Here I am defining $X$ is closed iff $T\setminus X$ is open.
 A: We say $x_n\to x$ iff for every open set $U\ni x$, almost all $x_n\in U$.
(Note that $x_n\to x$ and $x_n\to y$ need not imply $x=y$ in general!)
Let $X=\mathbb R$ with co-countable topology, i.e. $A$ is open iff $A=\emptyset$ or $\mathbb R\setminus A$ is at most countable. Then $T:=(-\infty,0]$ contains all its limit points because for any sequence $x_n$ in $T$ and point $y>0$, the set $\mathbb R\setminus\{x_n\mid n\in\mathbb N\}$ is an open neighbourhood of $y$ not containing any members of the sequence, i.e. $y>0$ is not limit of any sequence in $T$. But $T$ is not closed because it is not countable (and not all of $\mathbb R$).
The reason is (cf. ronno's comment) that $X$ is not first countable.
A: A subset $F$ of a topological space $X$ is called sequentially closed if every limit point $x$ of some sequence $(x_n)_{n\in\mathbb N}$ in $F$ is an element of $F$. A subset $U$ of $X$ is called sequentially open if every sequence $(x_n)$ converging to a point of $U$ is eventually in $U$. It is easy to show that the complement of a sequentially open set is sequentially closed, and a bit harder to prove the other direction.
A space $X$ is called sequential if every sequentially closed subset of $X$ is closed. The other implication is always true: A closed subset is always sequentially closed. For an example of a non-sequential space see Hagen's answer.
A space with the property that for every adherence point $x$ of a subset $A$ there is a sequence $(x_n)\subseteq A$ converging to $x$ is sequential, i.e. every first-countable space is sequential.
