Prove that a set $E$ is closed iff it's complement $E^{c}$ is open

I was wondering if this proof was right.

$\Leftarrow$ Suppose $E$ is closed. Then choose $x\in E^{c}$, then $x\notin E$, and so $x$ is not a limit point of $E$.

Hence there exists a neighborhood $N$ of $x$ such that $E \cap N$ is empty, such that $N \subset E^{c}$. Thus $x$ is an interior point of $E^{c}$ and $E^{c}$ is open

$\Rightarrow$ Then suppose $E^{c}$ is open. Let $x$ be a limit point of $E$, then every neighborhood of $x$ contains a point of $E$, so that $x$ is not an interior point of $E^{c}$. Since $E^{c}$ is open, this means that $ x \in E$, Therefore $E$ is closed.

Solution 1:

Looks good. All true statements and they make logical sense. However, it kinda depends on where you start. Is this meant to be a proof for metric spaces? Then it's absolutely fine. Or is it a proof for topological spaces? For topological spaces, the definition of being a closed set is usually that the complement is open. You could use the same definition for metric spaces, but that's usually not done. So it depends on the definitions from which you start.

Solution 2:

Yeah your proof looks fine, but you should give us the definition of closed and open. Because it is not unusual to define a set is closed when its complement is open. You should maybe defend what happens when $E=\varnothing$ or $E^C = \varnothing$.