Open sets and closed sets proof 
Theorem A set E is open if and only if the complement $E^{c}$ is closed

Proof
$E$ open $\iff$ any point $x \in E$ is an interior point
$\iff \forall x \in E, \exists$ a neighborhood $N$ of x s.t. $N$ is disjoint from $E^{c}$
$\iff \forall x\in E$, $x$ is not a limit point of $E^{c}$
$\iff E^{c}$ contains all its limit point


Here is my question for the last justification. How does showing that $x$ is not a limit point of $E^{c}$ imply that the $E^c$ contains all the limit points?
 A: Suppose that $E^c$ had a limit point $x$ that was not in $E^c$; then $x$ would belong to $E$ and therefore would not in fact be a limit point of $E^c$.
In other words, if nothing in $E$ is a limit point of $E^c$, then any limit points that $E^c$ might have can’t be in $E$ and must therefore be in $E^c$. (Remember, anything that’s not in $E$ is in $E^c$ and vice versa.)
A: If every $x\in E$ is not a limit point of $E^{c}$, then no limit point of $E^{c}$ can be found in $E$; otherwise, there would be some $x\in E$ which is a limit point of $E^{c}$, but we have already shown this to be false.  So, since no limit point of $E^{c}$ can be found in $E$, it follows that every limit point of $E^{c}$ is in $E^{c}$.  So, $E^{c}$ contains all of its limit points as desired and $E^{c}$ is closed.
A: All the limit points of $E^\rm{c}$ are either in $E^\rm{c}$ or in $E$. You've shown that points in $E$ cannot be limit points so that can only mean that $E^\rm{c}$ contains all it's limit points.
A: The following two statements are equivalent (they are each other's contrapositive):

If $x\in E$, then $x$ is not a limit point of $E^c$. (A rephrasing of $\forall x\in E$, $x$ is not a limit point of $E^c$.)
If $x$ is a limit point of $E^c$, then $x\notin E$ (that is $x\in E^c$).

That's how you make that last leap.
