Prove that the Complex plane is closed, open and perfect. Prove that the Complex plane is closed, open and perfect.
My intuition is destroyed by the fact that a set can be open and closed at the same time. 
The following is my understanding.
open: If all points in set $E$ is interior to $E$, then $E$ is open.
I think this means that all points $p$ in $E$ has a neighborhood that is a proper subset of $E$.
closed: If every limit point of $E$ is a point of $E$, then $E$ is closed.
I think this means that all neighborhoods of every limit point $p$ in $E$ contains a distinct point in $E$.
Perfect: If $E$ is closed and if every point of $E$ is a limit point of $E$.
I'm not quite sure I understand the difference between a point and a limit point...
This reminds me, since the complement of an open set is closed, does that mean that the complement of the complex plane, the empty set, is neither open nor closed ?
 A: The empty set and the whole space are open and closed at the same time. A limit point is a point that does not belong to the given set, but belongs to some super set (otherwise it makes no sense). The whole plane is open because every point is interior (it has no frontier). It is closed, because it contains all the points, in particular, the limit points. Finally, it is perfect, because any point is the limit point: Take any point, in any neighborhood of it, there are infinitely many points of the plane.
A: In fact, there are examples (even more radical - see below) which show that being open and closed at the same time is not impossible.
For now, I suggest you to think in terms of a definition of closedness in general topology space, not necessarily metric one. Here it goes.  
Definition A subset $F\subset X$ of a topology space $(X,\mathcal{T})$ is closed if its complement $F^c$ is open.
(If the space is metric, the definition above and that via limit points are equivalent.)
Now, what bothers you is a direct consequence of the fact that the whole space is always open, $X\in \mathcal{T}$, by the definition of topology. It suffices to note $\emptyset^c=X.$
Example. Let $X=\mathbb{N}$ be a set endowed with a discrete topology, i.e. such that every point is an open set. Since topology contains arbitrary unions of its own elements, $\mathcal{T}=2^{\mathbb{N}}$ follows.
Now it's strikingly clear that every set is closed as well as open (in fact it is clopen according to the standard terminology), as complement of a given singleton $\{n\}$ is closed.
A: In the topology $\tau_X$ on the given set $X$, every element of $\tau_X$ is open. $\emptyset$ and $X$ are both open and closed. However, any other element $U\in \tau_X$ need not be open and closed.

Example 1 Any element of the discrete topology on the set $X$ is open and closed.
Example 2 Any element( not $\emptyset$ or $\mathbb N$) of the finite complete topology on $\mathbb N$ cannot be open and closed.

