Limit point definition of a closed set, versus complement of open set definition? I have seen two definitions of closed sets:
1) the complement of a closed set is an open set
2) a closed set contains all its limit points
Are these two definitions essentially equivalent? Or is there some subtlety I might be missing? 
Where I am getting stuck is trying to figure out why an open set might not contain all of its limit points, in general. In particular, say that a set is $S$ is open: this means that every $V_\epsilon(x) \in S$. Let me put a further condition on $S$, stating that forall $d$, $V_d(x)$ contains at least two points---so every point is also a limit point. So the set $S$ is open and closed.
Okay, so clearly, "open" and "closed" are not complements of each other, since a set can be open and closed. This makes definition 1) difficult for me to swallow. Also, a set having all of its limit points is not not a property that excludes open sets, making definition 2) difficult to swallow.
Why not work with the following definition of closed/open:
The boundary points of a set $S$ are those where all $V_\epsilon(b) \cap S \neq V_\epsilon(b)$. A set is open if it contains no boundary points. The complement of such an open set is closed.
But there is some usefulness to way things are defined right now, and my definition probably has other issues I haven't yet figured out. There is a lot of subtlety I am clearly missing---that is one of the things I am beginning to realize as I play with topology.
 A: Suppose $C^c$ is open. Suppose $x $ is a limit point of $C$. If $x \notin C$ then there is some open $V$ such that $x \in V \subset C^c$, which contradicts
$x$ being a limit point.
Suppose $C$ contains all its limit points. Suppose $x \notin C$. Since $x$ is not
a limit point, there is some open $V$ such that $x \in V$ and $V$ does not intersect $C$. Hence $C^c$ is open.
A: I am answering the following question, which I think is the one you want answered:"might not an open set  contain all of its limit points, in general?" That's where you say you are getting stuck. You should get stuck there, because the answer is yes an open set can also be closed. A cheap shot is to take every subset of say the set of real numbers to be open; then every subset is also closed. Or take X to be the union of the open intervals (2,3) and (5,17). I'm hoping you will think of the topology on this set that I have in mind. If you do, you will see that (2,3) is open and contains all its limit points. A caveat: Although the definitions are equivalent, in topology definitions involving points are replaced by definitions involving open sets. Be on the lookout for this theme. 
